It takes 0.05 seconds for a part undergoing simple harmonic motion with a frequency of 5.00 Hz to move from x=0 to x=-1.80 cm.
Explanation:The student is asking how long it takes for a machine part undergoing simple harmonic motion (SHM) with a frequency of 5.00 Hz and amplitude of 1.80 cm to move from its equilibrium position (x=0) to its maximum negative displacement (x=-1.80 cm).
In SHM, the period (T) is the time required for one complete cycle of the motion. The period can be calculated using the formula T = 1/f, where f is the frequency. For a frequency of 5.00 Hz, the period is T = 1/5.00 Hz = 0.20 s. In half of a period, the part moves from one extreme to the other, passing through the equilibrium position at the quarter-period. Thus, to go from x=0 to x=-1.80 cm, it takes a quarter of this period, or (1/4)T = 0.05 s.
Therefore, it takes 0.05 seconds for the part to travel from x=0 to x=-1.80 cm while undergoing SHM.
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Two semiconductors are identical except that one has a band gap of 1.2 eV, while the other has a band gap of 1.1 eV. The room temperature (T = 300 K) intrinsic carrier density of the 1.1 eV material is ni1(300K) = 1.0×1019 m−3 and that of the 1.2 eV material is ni2(300K).1)Calculate the ratio of intrinsic carrier densities for the two materials at room temperature. ni2/ni1=7.56×10−21
To solve this problem it is necessary to apply the relationship given by the intrinsic carrier concentration, in each of the phases.
The intrinsic carrier concentration is the number of electrons in the conduction band or the number of holes in the valence band in intrinsic material. This number of carriers depends on the band gap of the material and on the temperature of the material.
In general, this can be written mathematically as
[tex]\eta_i = \sqrt{N_cN_v}e^{-\frac{E_g}{2KT}}[/tex]
Both are identical semiconductor but the difference is band gap which is:
[tex]E_{g1} = 1.1eV[/tex]
[tex]n_{i1} = 1*10^{19}m^{-3}[/tex]
[tex]E_{g2} = 1.2eV[/tex]
[tex]T=300K[/tex]
The ratio between the two phases are given as:
[tex]\frac{\eta_{i1}}{\eta_{i2}} = \frac{e^{-\frac{E_{g1}}{2KT}}}{e^{-\frac{E_{g2}}{2KT}}}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} = e^{\frac{E_{g2}-E_{g1}}{2KT}}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} =e^{\frac{(1.2-1.1)(1.6*10^{-19})}{2(1.38*10^{-23})(300)}}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} =e^{-1.932367}[/tex]
[tex]\frac{\eta_{i1}}{\eta_{i2}} =0.145[/tex]
Therefore the ratio of intrinsic carrier densities for the two materials at room temperature is 0.145
A satellite orbiting the moon very near the surface has a period of 110 min. Use this information, together with the radius of the moon r=1.74×106m, to calculate the free-fall acceleration on the moon’s surface. What is the moon's acceleration due to gravity?
To solve this problem it is necessary to apply the concepts related to the centripetal Force, the Force of the weight produced by the gravity of the star and the definition of the velocity as a function of time (period)
In other words, for balance to exist, the force acting on the body (weight) must be equal to the centripetal force that attracts it in motion. That is to say
[tex]\sum F = 0[/tex]
[tex]mg_{moon}-ma_r=0[/tex]
[tex]mg_{moon}=ma_r[/tex]
[tex]g_{moon} = \frac{v^2_{sat}}{r}[/tex]
The value of the given period is 110min or 6600s.
At the same time we know that the speed of a body depending on its period (simple harmonic movement) is subject to
[tex]v_{sat} = \frac{2\pi r}{T}[/tex]
[tex]v_{sat} = \frac{2\pi (1.74*10^6)}{6600s}[/tex]
[tex]v_{sat} = 1.66*10^3m/s[/tex]
Finally, replacing this value in the event of gravity we have to
[tex]g_{moon} = \frac{v^2_{sat}}{r}[/tex]
[tex]g_{moon} = \frac{(1.66*10^3)^2}{1.74*10^6}[/tex]
[tex]g_{moon} = 1.58m/s^2[/tex]
What can you conclude about the nature of vertical acceleration for a freely falling projectile?
Answer:
For a free falling projectile the acceleration of the body is always equal to the acceleration due to gravity and in the downward direction.
Explanation:
The acceleration of a body falling freely will experience the acceleration always in the downward direction.When the height of free fall is of the order of radius of the earth then we have a variable acceleration with respect to its height. This is given as:[tex]a=g.\frac{R^2}{(R+h)^2}[/tex]
where:
g = acceleration due to gravity on the earth surface
R = radius of the earth
h = height of the body above the earth surface
Final answer:
In a freely falling projectile, vertical acceleration is constant at 9.8 m/s² due to gravity, with velocity decreasing on the ascent and increasing on the descent, independent of horizontal motion.
Explanation:
Understanding Vertical Acceleration in Freely Falling Projectiles
When examining the characteristics of a freely falling projectile, we can make specific conclusions about its vertical acceleration. This acceleration is caused by the force of gravity, and its value is constant at 9.8 m/s² in the downward direction, irrespective of any horizontal motion the object may have.
This means that during the upward flight, the magnitude of the vertical velocity decreases, becoming zero at the projectile's highest point. Once the projectile starts its descent, the velocity increases again at the same rate due to the consistent force of gravity acting on it vertically.
The unique aspect of projectile motion is the independence of vertical motion from horizontal motion. Despite any horizontal velocity, the vertical acceleration remains unaffected and maintains a steady value of 9.8 m/s². This is a fundamental concept that was of great interest historically for both scientific and military applications and still holds significant relevance in physics.
A stationary police car emits a sound of frequency 1240 Hz that bounces off of a car on the highway and returns with a frequency of 1265Hz . The police car is right next to the highway, so the moving car is traveling directly toward or away from it. And what frequency would the police car have received if it had been traveling toward the other car at 25.0 m/s?
Answer:
1467.56 Hz
Explanation:
given,
f₀ = 1215 Hz,
f₂ = 1265 Hz
speed of sound u = 340 m/s,
speed of the other car = v
when the police car is stationary:
the frequency the other car receives is
[tex]f_1= f_0\times \dfrac{u+v}{u}[/tex]
the frequency the police car receives is
[tex]f_2 = f_1\times \dfrac{u}{u-v}[/tex]
now,
[tex]\dfrac{f_2}{f_0}=\dfrac{u+v}{u-v}[/tex]
[tex]\dfrac{1265}{1240}=\dfrac{u+v}{u-v}[/tex]
[tex]v =\dfrac{1265-1240}{1265+1240}\times u[/tex]
[tex]v =\dfrac{1265-1240}{1265+1240}\times 344[/tex]
v = 3.43 m/s
when the police car is moving with v_p = 25.0 m/s toward the other car:
the frequency the other car receives is
[tex]f_1= f_0\times \dfrac{u+v}{u-v_p}[/tex]
the frequency the police car receives is
[tex]f_2 = f_1\times \dfrac{u+v_p}{u - v}[/tex]
now,
[tex]= f_0\times \dfrac{(u+v)(u + v_p)}{(u-v)(u-v_p)}[/tex]
[tex]=1240\times \dfrac{(344 +3.43)(344 + 25)}{(344-3.43)(344 - 25)}[/tex]
= 1467.56 Hz
Using the Doppler Effect formula, where f' = f(v + vo) / v, one could calculate the frequency the police car would have received if it had been traveling toward the oncoming car at 25.0 m/s. This calculation would take into account the initial Doppler shift from the moving car that was recorded by the stationary police car and adjust for the police car's new velocity.
Explanation:The question involves the concept of the Doppler Effect in physics, specifically in the context of sound waves. To find the frequency the police car would have received if it had been traveling toward the other car at 25.0 m/s, the Doppler Effect formula for frequency observed by a moving observer and stationary source can be used:
f' = f(v + vo) / (v)
where f is the source frequency, f' is the observed frequency, v is the speed of sound, and vo is the velocity of the observer (police car in this case), with direction toward the source being positive. Given the original frequencies (1240 Hz emitted and 1265 Hz received), we can infer the car was moving toward the police car to cause this increase in frequency.
Using the formula with the given values and assuming a standard speed of sound at room temperature, which is typically around 343 m/s, we can calculate the received frequency. In this case, we will also need to reverse the initial Doppler shift effect from the moving car before applying the observer's (police car's) motion effect. This involves a complex calculation where we first find the car's velocity using the initial frequencies and then solve for the new frequency with the police car moving.
A violinist tuning her violin plays her A-string while sounding a tuning fork at A 415 Hz, and hears 5 beats per second. When she tightens the string, the beat frequency decreases. What should she do to tune the string to A, and what was the original untuned freq of her string
Answer:410 Hz
Explanation:
It is given that frequency of tuning fork is 415 Hz and hears a beat of 5 beat per second
so untuned frequency must be either 410 Hz or 420 Hz because beat frequency is the difference in frequency of two notes.
When she tightens the string the beat frequency decrease
and velocity [tex]v=\nu \lambda [/tex]
where [tex]\nu =frequency[/tex]
[tex]\lambda =wavelength[/tex]
[tex]v=velocity[/tex]
[tex]v=\sqrt{\frac{T}{\mu }}[/tex]
where T=tension
it T increase v also increases and from 1 st equation frequency is also increasing therefore untuned frequency must be 410 Hz because beat frequency is decreasing.
A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface.
1) At what point in the cycle does the penny first lose contact with the piston? (answer choices:)
A. midpoint moving up
B. midpoint losing down
C. highest point
D. lowest point
2) What is the maximum frequency for which the penny just barely remains in place for the full cycle?
1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.
2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as
[tex]a = -\omega^2 A[/tex]
Where,
a = Acceleration
A = Amplitude
[tex]\omega[/tex]= Angular velocity
From a reference system in which the downward acceleration is negative due to the force of gravity we will have to
[tex]a = -g[/tex]
[tex]-\omega^2 A = -g[/tex]
[tex]\omega = \sqrt{\frac{g}{A}}[/tex]
From the definition of frequency and angular velocity we have to
[tex]\omega = 2\pi f[/tex]
[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}[/tex]
[tex]f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}[/tex]
[tex]f = 2.5Hz[/tex]
Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz
Final answer:
The penny loses contact with the piston at the highest point in the cycle, and the maximum frequency for which the penny remains in place is at its lowest point.
Explanation:
In simple harmonic motion, the penny loses contact with the piston at the highest point in the cycle. This is because at the highest point, the acceleration due to gravity is greater than the restoring force provided by the piston. Therefore, the correct answer is C. highest point.
The maximum frequency for which the penny just barely remains in place for the full cycle is when the acceleration due to gravity is equal to the restoring force provided by the piston. At this frequency, the penny is in equilibrium and does not leave the surface. Therefore, the maximum frequency is when the penny is at its lowest point in the cycle. So, the correct answer is D. lowest point.
The magnitude of the magnetic field in a magnetic resonance imaging (MRI) machine can be as great as B = 2.0 T . Under normal circumstances, this field cannot be shut off by just flipping a switch. Instead the magnitude needs to be carefully decreased to zero. In an emergency, however, the magnet can be "quenched" so that B reduces to zero in 20 s. Such a quench can cost thousands of dollars and likely damages the magnets. Assume that the magnetic field exists inside a cylinder of radius R = 400 mm and length ℓ = 300 mm .
1. How much magnetic potential energy is dissipated when the magnetic field is quenched in this way?2. What is the average rate at which energy is dissipated?
Answer
given,
B = 2 T
time = 20 s
Radius = R = 400 mm
L= 300 mm
apply the formula for magnetic energy density
[tex]\eta_0 = \dfrac{energy}{volume}[/tex]
[tex]\eta_0 = \dfrac{B^2}{2\mu}[/tex]
[tex]\eta_0 = \dfrac{2^2}{2\times 4 \pi 10^{-7}}[/tex]
energy density = 1.59 x 10⁶ J/m³
Now , in the cylinder ,
Energy = energy density x volume
E = 1.59 x 10⁶ x π x r² x h
E = 1.59 x 10⁶ x π x 0.4² x 0.5
E = 4 x 10⁵ J
now, average rate of energy dissipated
rate of dissipation of energy = Energy/time
= 4 x 10⁵ /20
= 2 x 10⁴ W
the average rate at which energy is dissipated is 2 x 10⁴ W
1. The energy be "4 × 10⁵ J".
2. The average rate be "2 × 10⁴ W"
Magnetic fieldAccording to the question,
Time, t = 20 s
Radius, R = 400 mm
Length, L = 300 mm
We know the formula,
Magnetic energy density,
→ [tex]\eta_0[/tex] = [tex]\frac{Energy}{Volume}[/tex]
= [tex]\frac{B^2}{2 \mu}[/tex]
By substituting the values, we get
= [tex]\frac{(2)^2}{2\times 4 \pi 10^{-7}}[/tex]
= 1.59 × 10⁶ J/m³
Now,
1.
Energy will be:
= Energy density × Volume
= 1.59 × 10⁶ × π × (0.4)² × 0.5
= 4 × 10⁵ J
2.
We know,
The rate of dissipation = [tex]\frac{Energy}{Time}[/tex]
= [tex]\frac{4\times 10^5}{20}[/tex]
= 2 × 10⁴ W
Thus the above response is correct.
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Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to the surface with lungs full of air, his lungs could be damaged. Explain why this is so.
Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).
This means the lungs are inflated with - Highly pressurized- gas.
This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.
If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.
The gas in the diver's lungs will - expand - and can damage the alveoli.
A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a particular design. The tubular shaft has an inner radius of ci = 0.300 in The shafts are to be powered by a motor operating at a frequency of f = 2.70 Hz and are to handle an attached load. Assume the cross sections are uniform throughout the lengths of the shafts and that the materials have an allowable shear stress of τallow = 14.5 ksi .
Answer:
The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp
Explanation:
Polar moment of Inertia
[tex](I_p)s = \frac{\pi(0.55)4}2[/tex]
= 0.14374 in 4
Maximum sustainable torque on the solid circular shaft
[tex]T_{max} = T_{allow} \frac{I_p}{r}[/tex]
=[tex](14 \times 10^3) \times (\frac{0.14374}{0.55})[/tex]
= 3658.836 lb.in
= [tex]\frac{3658.836}{12}[/tex] lb.ft
= 304.9 lb.ft
Maximum sustainable torque on the tubular shaft
[tex]T_{max} = T_{allow}( \frac{Ip}{r})[/tex]
= [tex](14 \times10^3) \times ( \frac{0.13101}{0.55})[/tex]
= 3334.8 lb.in
= [tex](\frac{3334.8}{12} )[/tex] lb.ft
= 277.9 lb.ft
Maximum sustainable power in the solid circular shaft
[tex]P_{max} = 2 \pi f_T[/tex]
= [tex]2\pi(2.1) \times 304.9[/tex]
= 4023.061 lb. ft/s
= [tex](\frac{4023.061}{550})[/tex] hp
= 7.315 hp
Maximum sustainable power in the tubular shaft
[tex]P _{max,t} = 2\pi f_T[/tex]
= [tex]2\pi(2.1) \times 277.9[/tex]
= 3666.804 lb.ft /s
= [tex](\frac{3666.804}{550})[/tex]hp
= 6.667 hp
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at 38.0 cm/s. Find its (a) angular frequency and (b) wave number. (c) Write a mathematical expression describing the displacement y of this wave (in centimeters) as a function of position and time. Assume the maximum displacement occurs when t = 0.
Answer
given,
wavelength = λ = 18.7 cm
= 0.187 m
amplitude , A = 2.34 cm
v = 0.38 m/s
A) angular frequency = ?
[tex]f = \dfrac{v}{\lambda}[/tex]
[tex]f = \dfrac{0.38}{0.187}[/tex]
[tex]f =2.03\ Hz[/tex]
angular frequency ,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) the wave number ,
[tex]K = \dfrac{2\pi}{\lambda}[/tex]
[tex]K= \dfrac{2\pi}{0.187}[/tex]
[tex]K =33.59\ m^{-1}[/tex]
C)
as the wave is propagating in -x direction, the sign is positive between x and t
y ( x ,t) = A sin(k x - ω t)
y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)
The angular frequency is 12.72 rad/s, the wave number is 0.34 cm^-1, and the displacement y of the wave as a function of position x (in cm) and time t (in seconds) is given by y = 2.34 sin(0.34x + 12.72t).
To answer this question, we need to address it part by part.
(a) The angular frequency (ω) of a wave is related to its speed (v) and wavelength (λ) by the formula ω = 2πv/λ. Substituting the given numbers, ω = 2π(38.0 cm/s)/(18.7 cm) = 12.72 rad/s.(b) Next comes the wave number (k). It is related to the wavelength by the formula k = 2π/λ. So, substituting value of λ, k = 2π/(18.7 cm) = 0.34 cm^-1.(c) The equation governing the displacement y (in cm) of a simple harmonic wave moving in the negative x direction, with maximum displacement at t = 0, is given by y = A sin(kx + ωt). In this case, A is the amplitude (2.34 cm), k is the wave number, and ω is the angular frequency. Hence, the equation is y = 2.34 sin(0.34x + 12.72t).It is important to note that all the calculations are done in the commonly used SI system of measurements.
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The tape in a videotape cassette has a total length 191 m and can play for 1.9 h. As the tape starts to play, the full reel has an outerradius of 35 mm and an inner radius of 11 mm. At some point during the play, both reels will have the same angular speed.What is this common angular speed?
The common angular speed is 9090.91 rad/h.
Explanation:To find the common angular speed, we need to first find the initial and final radii of the full reel. The initial radius is 35 mm and the final radius is 11 mm. The angular speed can be found using the equation v = rω, where v is the linear speed, r is the radius, and ω is the angular speed. We can solve for ω by dividing the linear speed by the radius. Since both reels will have the same angular speed, we can use the same equation to find the angular speed when the tape has unspooled a distance of 191 m.
First, let's convert the radii from millimeters to meters:
Initial radius: 35 mm = 0.035 mFinal radius: 11 mm = 0.011 mNext, we'll find the linear speeds at the initial and final radii. Since the length of the tape is 191 m and it plays for 1.9 hours, the linear speed can be calculated by dividing the length of the tape by the playtime:
Linear speed at initial radius: (191 m)/(1.9 h) = 100 m/hLinear speed at final radius: (191 m)/(1.9 h) = 100 m/hFinally, we can find the common angular speed by dividing the linear speed by the radius:
Initial angular speed: (100 m/h)/(0.035 m) = 2857.14 rad/hFinal angular speed: (100 m/h)/(0.011 m) = 9090.91 rad/hTherefore, the common angular speed is 9090.91 rad/h.
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A web page designer creates an animation in which a dot on a computer screen has position ⃗r=[4 cm+(2.5cm/s 2 )t 2 ] ^ i+(5cm/s)t ^ j .
a) Find the magnitude and direction of the dot’s average velocity between t=0 and t=2 s.
b) Find the magnitude and direction of the instantaneous velocity at t=0, t=1 s, and t=2 s. and show the velocities calculated in pan (b).
Answer:
Explanation:
a )
r = (4 + 2.5 t² )i + 5 t j
When t = 0
r₁ = 4 cm i
When t = 2s
r₂ = 14 i + 10 j
Displacement
= r₂ - r₁
=14 i + 10 j -4i
= 10i + 10j
average velocity
= displacement / time
=( 10i + 10j )/2
= 5i + 5j
b )
r = (4 + 2.5 t² )i + 5 t j
dr / dt = 5 t i + 5 j
Instantaneous velocity at t = o
(dr / dt)₀ = 5 j
Instantaneous velocity at t = 1s
(dr / dt)₁ = 5i + 5j
Instantaneous velocity at t = 2s
(dr / dt)₂ = 10i + 5 j
(a) The change in position or displacement (x) divided by the time intervals (t) in which the displacement happens is the average velocity. the magnitude and direction of the dot’s average velocity between t=0 and t=2 s will be 5 m/sec, 5 m/sec.
(b) the velocity of an object at a given point in time. . the magnitude and direction of the instantaneous velocity at t=0, t=1 s, and t=2 s. will be 10 m/sec and 5 m/sec.
(a) What is average velocity?The change in position or displacement (x) divided by the time intervals (t) in which the displacement happens is the average velocity
given,
[tex]\vec{r}= (4+2.5t^{2})\hat{i}+5t\hat{j}[/tex]
At time t=0
[tex]r_1= 4 \hat{i}[/tex][tex]\vector{r_2} -\vector{r_1}= 14 \hat{i}+10\hat{j}-4\hat{i}[/tex]
at (t=2)
[tex]\vector{r_2}= 14 \hat{i}+10\hat{j}[/tex]
[tex]\vector{r_2} -\vector{r_1}= 10 \hat{i}+10\hat{j}[/tex]
Average velocity = [tex]\frac{\vector{r_2} -\vector{r_1}}{t_2-t_1}[/tex]
Average velocity =[tex]5\hat{i}+5 \hat{j}[/tex]
hence the magnitude and direction of the dot’s average velocity between t=0 and t=2 s will be 5 m/sec, 5 m/sec.
(b)What is instantaneous velocity?instantaneous velocity is the velocity of an item in motion at a single point in time.
Given,
[tex]\vec{r}= (4+2.5t^{2})\hat{i}+5t\hat{j}[/tex]
[tex](\frac{ \del{r}}{\del{t}})_0= 5\hat{j}[/tex][tex](\frac{ \del{r}}{\del{t}})_2= 10\hat{i}+5\hat{j}[/tex]
[tex](\frac{ \del{r}}{\del{t}})_1= 5\hat{j}+5\hat{J}[/tex]
Hence the magnitude and direction of the instantaneous velocity at t=0, t=1 s, and t=2 s. will be 10 m/sec and 5 m/sec.
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You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower
Answer:
a) L = 0.75m f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m f₁ = 56.67 Hz , f₃ = 170 Hz
Explanation:
This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung
In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength
With 1 node λ₁ = 4 L
With 2 nodes λ₂ = 4L / 3
With 3 nodes λ₃ = 4L / 5
The general term would be λ_n= 4L / n n = 1, 3, 5, ((2n + 1)
The speed of sound is
v = λ f
f = v / λ
f = v n / 4L
Let's consider each length independently
L = 0.75 m
f₁ = 340 1/4 0.75 = 113.33 n
f₁ = 113.33 Hz
f₃ = 113.33 3
f₃ = 340 Hz
L = 1.5 m
f₁ = 340 n / 4 1.5 = 56.67 n
f₁ = 56.67 Hz
f₃ = 56.67 3
f₃ = 170 Hz
The lowest two frequencies of resonance for the shower enclosure are 229 Hz and 114.33 Hz. The second harmonics for these lengths would be 458 Hz and 228.67 Hz respectively.
Explanation:The lowest two frequencies that correspond to resonances in your shower enclosure, which is 0.75 m wide and 1.5 m long, can be found using the formula for the fundamental frequency of a standing wave: f = v/2L, where v is the speed of sound in the air (~343 m/s), and L is the length or width of the enclosure, respectively.
For the axis parallel to the width (0.75 m), the lowest frequency is f = 343/(2*0.75) = 229 Hz. For the second harmonic along this axis, the frequency would be twice the fundamental, so 2*229 = 458 Hz.
For the axis parallel to the length (1.5 m), the lowest frequency is f = 343/(2*1.5) = 114.33 Hz. For the second harmonic along this axis, the frequency would be twice the fundamental, so 2*114.33 = 228.67 Hz.
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Which of the following is most responsible for the magnetic behavior of materials?
A) neutronsB) protonsC) photonsD) electronsE) phonons
Answer:
D) electrons
Explanation:
Moving charges creates magnetic field. So the moving electrons of an atom generate microscopic magnetic field. In an atom the magnetism is destroyed when it has equal no. of electrons spinning in the opposite direction.
In the atoms of iron, cobalt and nickle there are more no. of electrons spinning in the same direction.
Electrons are the primary contributors to the magnetic behavior of materials due to their spin, making them responsible for magnetism in materials.
Explanation:The element most responsible for the magnetic behavior of materials is electrons. This is because electrons, as well as protons, possess an intrinsic magnetic property known as spin, which plays a significant role in magnetism. While neutrons also have a magnetic field, they do not contribute significantly to the magnetic properties of materials as they have no net electric charge. Therefore, the correct answer to the student's question is D) electrons.
A puddle holds 150 g of water. If 0.50 g of water evaporates from the surface, what is the approximate temperature change of the remaining water? (Lv = 540 cal/g)
a. +1.8 C°
b. -1.8 C°
c. +0.18 C°
d. −0.18 C°
Answer:
The correct answer is b
Explanation:
In this exercise we must assume that no heat from the environment enters, the initial heat of the puddle is distributed in the heat of the evaporated water and the heat of liquid water remaining
Let's look for the heat to evaporate the water
Q₁ = m L
Q₁ = 0.50 540
Q₁ = 270 cal
The remaining water is
m = 150 -0.50 = 149.5 gr
The heat for this water is
Q2 = m ce DT
The amount of heat should be conserved, the heat assigned is equal to minus the heat absorbed
Q₁ = - Q₂
Q₁ = -m ce ΔT
ΔT = -Q₁ / m ce
Δt = -270 / (149.5 1)
ΔT = -1.8 ° C
The water has cooled 1.8 ° c
The correct answer is b
As part of a science project you investigate the effect of punching holes in a plastic container of water and observing the flow of liquid out of the different holes. If the container has a water level 20 cm above the base and you punch a hole in the side 12 cm high, calculate the speed at which water will initially flow out of the hole.
a.1.25 m/s
b.15.3 m/s
c.12.5 m/s
d. 1.50 m/s
Final answer:
Using Bernoulli's principle and Torricelli's law, the initial speed of water flowing out of a hole in a container can be calculated. The speed is determined by the height of water above the hole and the acceleration due to gravity, resulting in an initial speed of approximately 1.25 m/s for the given problem.
Explanation:
The question involves calculating the initial speed at which water will flow out of a hole in a plastic container, using principles of fluid dynamics found in physics. To determine this speed, Bernoulli's principle and Torricelli's law can be applied. These principles state that the speed (v) of the fluid exiting the hole can be calculated using the equation v = √(2gh), where g is the acceleration due to gravity (9.81 m/s²) and h is the height of the water column above the hole.
In this case, h = 20 cm - 12 cm = 8 cm = 0.08 m. Plugging this into the equation gives v = √(2 * 9.81 m/s² * 0.08 m), which results in v ≈ 1.25 m/s. Therefore, the initial speed of the water flowing out of the hole is approximately 1.25 m/s, making option a the correct answer.
The current in an oscillating LC circuit is zero at a particular moment.
Which one of the following statements is true?
A) The charge on the capacitor is equal to zero coulombs.
B) Charge is moving through the inductor.
C) The energy is equally shared between the electric and magnetic fields.
D) The energy in the electric field is maximized.
E) The energy in the magnetic field is maximized.
Answer:
The current in an oscillating LC circuit is zero when the charge on the capacitor is equal to zero coulombs.
Explanation:
Because when the charge would not flow through the circuit then current in an oscillating circuit will be zero.
In an oscillating LC circuit, when the current is at zero, it implies that the system is fully charged or discharged, which means the energy in the electric field is maximized.
Explanation:An LC circuit, also known as a resonant circuit or a tuned circuit, is a type of electronic circuit that is composed of an inductor L and a capacitor C. Electric fields store charge in the capacitor, while magnetic fields store energy in the inductor. In reference to the question, if we know the current in an oscillating LC circuit is zero at a particular moment, then this implies that the charge is at its maximum or minimum point because the system is fully charged or discharged. The correct answer to the question is, therefore D) The energy in the electric field is maximized.
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While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 21.3days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.
Part A
Calculate the distance between the stars.
Express your answer to two significant digits and include the appropriate units.
d =
Answer:
Explanation:
Convert orbital period into seconds.
21.3 days = 21.3x 24 x 60x60 = 1840320 s
Write the expression for the gravitational force balanced by the linear speed to calculate the distance between the stars.
[tex]\frac{Gm^2}{d^2}=\frac{mv^2}{r}[/tex] [tex]d=2r;v=\frac{2\pi r}{T}[/tex]
[tex]\frac{Gm^2}{(2r)^2}=\frac{m(\frac{2\pi r}{T})^2}{r}\\\\\frac{Gm^2}{4r^2}=\frac{m(\frac{4\pi^2 r^2}{T^2})}{r}\\\\r^3=(\frac{GmT^2}{16\pi^2})[/tex]
[tex]r=(\frac{(6.67\times 10^{-11})(1.99\times 10^{30})(1840320)^2}{16(3.14)^2})^{\frac{1}{3}}\\\\=1.42\times 10^{10}m[/tex]
Calculate the distance between the stars.
[tex]d=2r\\\\=2(1.42\times 10^{10}m)\\\\=2.83\times 10^{10}m[/tex]
The distance between the two stars in the binary system can be calculated using Kepler's third law as approximately 0.150 Astronomical Units (AU).
Explanation:In physics, we can determine the distance between the stars using Kepler's third law which relates the orbital period of an object in a binary system to the distance between the two objects. The law is expressed as T² = R³ where T is the orbital period in years and R is the separation distance in Astronomical Units (AU). So first, we'll need to convert the given period of 21.3 days into years, which gives us approximately 0.058 years. Plugging the orbital period into Kepler's third law gives us a radius cubed of (0.058)² = 0.003364. Taking the cube root of both sides gives us a separation of about 0.150 AU between the stars.
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What general rule can you write to explain the forces between objects of different mass?
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects. if u look at the definition of force u get "strength or energy as an attribute of physical action or movement." and when you are talking about forces between two objects of a different mass u think of Force equals mass times acceleration [ ]. For every action there is an equal and opposite reaction. Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.
The general rule explaining forces between different masses is Newton's Universal Law of Gravitation. According to this, the gravitational pull between two objects is inversely proportional to the square of their distance apart and immediately relates to their masses.
Explanation:The general rule that explains the forces between objects of different masses is called Newton's Universal Law of Gravitation. According to this law, the gravitational pull between two bodies is inversely proportional to the square of their distance apart and immediately relates to the product of their masses. In general, an object's gravitational attraction is stronger the more mass it has. Essentially, the greater the mass of an object, the stronger its gravitational pull. Conversely, the farther apart two objects are, the weaker the gravitational pull between them.
The formula for this law is given as Fgravity = G * (M₁ M₂ / R²). Here, Fgravity represents the gravitational force, G is the gravitational constant, M₁ and M₂ are the masses of the two bodies, and R² is the square of the distance between the centres of the two bodies.
This law is fundamental to physics as it explains a huge range of phenomena, from the motion of planets in our solar system to the existence of tides on Earth.
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A satellite is in a circular orbit around the Earth at an altitude of 2.30 106 m. (a) Find the period of the orbit. 2.23 Correct: Your answer is correct. h (b) Find the speed of the satellite. 6.78 Correct: Your answer is correct. km/s (c) Find the acceleration of the satellite. 5.3 Correct: Your answer is correct. m/s2 toward the center of the Earth Need Help? Read It
Answer:
(a) T = 2.23h
(b) v = 6.78 km/s
(c) a = 5.31 m/s²
Explanation:
(a) The period (T) of the orbit can be calculated using Kepler's Third Law equation:
[tex] T = 2\pi \sqrt \frac{r^{3}}{GM} [/tex] (1)
where r: is the distance from the center of the Earth and the satellite, G: is the gravitational constant = 6.67x10⁻¹¹m³kg⁻¹s⁻² and M: is the Earth's mass = 5.97x10²⁴kg
Since [tex] r = r_{E} + r_{S} = 6.36\cdot 10^{6}m + 2.30\cdot 10^{6}m = 8.66\cdot 10^{6}m [/tex]
where [tex]r_{E}[/tex]: is the Earth's radius and [tex]r_{S}[/tex]: is the distance between the surface of the Earth and the satellite
Hence, by entering the radius calculated into equation (1) we can find the period of the orbit:
[tex] T = 2\pi \sqrt \frac{(8.66\cdot 10^{6}m)^{3}}{6.67\cdot 10^{-11}m^{3}/kgs^{2}\cdot 5.97\cdot 10^{24}kg} = 8024.3s = 2.23h [/tex]
(b) The speed (v) of the satellite can be calculated using the following equation:
[tex] v = \sqrt \frac{GM}{r} = \sqrt \frac{6.67\cdot 10^{-11}m^{3}/kgs^{2} \cdot 5.97\cdot 10^{24}kg}{8.66 \cdot 10^{6}m} = 6.78km/s [/tex]
(c) The acceleration (a) of the satellite is:
[tex] a = \frac{v^{2}}{r} = \frac{(6.78 \cdot 10^{3}m/s)^{2}}{8.66\cdot 10^{6}m} = 5.31m/s^{2} [/tex]
I hope it helps you!
The question involves calculating satellite orbital characteristics like period, speed, and acceleration, which can be done using Kepler's laws and formulas for centripetal force.
Explanation:The student's question revolves around the calculation of various orbital characteristics of a satellite, such as period, speed, and acceleration. Kepler's laws, particularly the third law, are essential to determine the period of an object in orbit. The period refers to the time it takes for a satellite to complete one full orbit around Earth. The speed of a satellite in orbit is computed by dividing the circumference of the orbit by the period, which provides an understanding of how fast the satellite travels through space. Lastly, the centripetal acceleration of a satellite is calculated using the formula for centripetal force, which depends on the mass of the satellite, its speed, and the radius of orbit. These calculations are fundamental for understanding satellite orbits and the forces that govern them.
Starting from rest at the origin, x0 = 0, a car accelerates in a straight line along the +x direction. For the time interval from t0 = 0 s to t1= 4 s, its velocity is given by v(t) = (−10 m/s2 )t + (6 m/s3)t2 . Then from t1 = 4 s to t2 = 10 s, it maintains a constant velocity v = 56 m/s in the same direction.
(i) Find the average acceleration for the first time interval, from t0 = 0 s to t1 = 4 s.
(ii) Find the average acceleration for the second time interval, from t1 = 4 s to t2 = 10 s
(iii) Find the instantaneous acceleration a(t) as a function of time for 0 s < t < 4 s.
(iv) Find the position x(t) as a function of time for 0 s < t < 4 s
Answer:
i) The average acceleration for the first time interval, from [tex]t_{0} = 0[/tex] to [tex]t_{1} =4\ s[/tex], is [tex]14 \frac{m}{s^{2} }[/tex] .
ii) The average acceleration for the second time interval, from [tex]t_{1} =4\ s[/tex] to [tex]t_{2} =10\ s[/tex], is zero.
iii) The instantaneous acceleration as a function of time a(t) between 0 s and 4 s is [tex]a(t)=-10\frac{m}{s^{2} } + 12\frac{m}{s^{3} } t[/tex] .
iv) The position as a function of time x(t) between 0 s and 4 s is [tex]x(t)=-5\frac{m}{s^{2} } t^{2} + 2\frac{m}{s^{3} } t^{3}[/tex] .
Explanation:
We are given as data that a car that is a rest, that means v₀=0, at the origin, x₀=0, accelerates in a straight line along the +x direction. We are told that from [tex]t_{0} = 0[/tex] to [tex]t_{1} =4\ s[/tex] the velocity is [tex]v(t)=-10\frac{m}{s^{2} } t + 6 \frac{m}{s^{3} } t^{2}[/tex] and that from [tex]t_{1} =4\ s[/tex] to [tex]t_{2} =10\ s[/tex] it maintains a constant velocity of [tex]v=56\frac{m}{s}[/tex] in the same direction.
i) The average acceleration is simply the rate of change of velocity, it can be expressed as:
[tex]\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0} }{t_{f}-t_{0}}[/tex]
we have to replace the values in this equation, the only value we need to calculate is [tex]v_{f}[/tex]
[tex]v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } (4 s)^{2}[/tex]
[tex]v(4 s)=-10\frac{m}{s^{2} } 4 s + 6 \frac{m}{s^{3} } 16 s^{2}[/tex]
[tex]v(4 s)= -40 \frac{m}{s} + 96 \frac{m}{s}= 56 \frac{m}{s}[/tex]
we get that [tex]v_{f}= 56 \frac{m}{s}[/tex] .
Then we pur this value in the expression of the average acceleration
[tex]\bar{a}=\frac{\vartriangle v}{\vartriangle t} =\frac{v_{f}-v_{0}}{t_{f}-t_{0}}=\frac{56\frac{m}{s}-0\frac{m}{s} }{4 s - 0 s}[/tex]
the average acceleration for the first time interval is
[tex]\bar{a}=14\frac{m}{s^{2} }[/tex]
ii) The average acceleration for the second time interval, from [tex]t_{1} =4\ s[/tex] to [tex]t_{2} =10\ s[/tex], is zero because the velocity is constant. The average acceleration is the rate of change of velocity, if this magnitude remains constant then it follows that the acceleration is zero.
iii) We calculate the instantaneous acceleration, which is the acceleration at a specific moment in time, as the derivative of the velocity function.
Mathematically [tex]a(t)=\frac{d v(t)}{d t }[/tex]
so we get
[tex]a(t)=-10 \frac{m}{s^{2} } + 12\frac{m}{s^{3} } t[/tex]
iv) To find the position in function of time x(t) we simply integrate the expression of v(t):
[tex]v(t)=\frac{d x(t)}{d t}= -10\frac{m}{s^{2} }t+ 6\frac{m}{s^{3} } t^{2}[/tex]
we get that
[tex]x(t)=-5\frac{m}{s^{2} } t^{2} + 2 \frac{m}{s^{3} } t^{2}[/tex]
Ifthe amount of ice on the planet increased to the amount that was present during the last glacial maximum, how muchwould it cause global sea level to fall? If you do not remember the approximate size of this rise, the volume of additionalice is 51000000 km3 and the surface area of the world's ocean is 361,000,000 km;A. less than 10 centimetersB. 10 centimeters to 1 meterC. 1 meter to 10 metersD. 10 meters to 50 metersE. more than 50 meters
Answer:
h = 12.95 m , The correct answer is D
Explanation:
For this exercise we use the definition of density
ρ= m V
For the water
[tex]\rho_{w}[/tex] = m / [tex]V_{w}[/tex]
For ice
[tex]\rho_{i}[/tex] = m / [tex]V_{i}[/tex]
The tabulated water density is [tex]\rho_{w}[/tex] = 997 Kg / m³ average between temperatures, the density near the freezing point that is 1000 kg/m³ can also be used, let's use the latter; The density of ice is 916.8 kg / m³
The mass of the water that is frozen is equal to the mass of ice that is formed, therefore, we can clear the doughs in the two equals formulas
Water
m = [tex]\rho_{w}[/tex] [tex]V_{w}[/tex]
Ice
m = [tex]\rho_{i}[/tex] [tex]V_{w}[/tex]
[tex]\rho_{w}[/tex] Vw = [tex]\rho_{i}[/tex] [tex]V_{i}[/tex]
Vw = [tex]\rho_{i}[/tex] / [tex]\rho_{w}[/tex] [tex]V_{i}[/tex]
Reduce to SI system
[tex]V_{i}[/tex] = 5.1 107 km³ (10³m / 1 km)³ = 5.1 10¹⁶ m³
Let's calculate
[tex]V_{w}[/tex] = 916.8 / 1000 5.1 10¹⁶
[tex]V_{w}[/tex] = 4,676 10¹⁶ m³
It indicates the surface of the ocean, so it is volume
V = A h
Where A is the surface and h is the height
A = 3.61 109 km² (10³ m / 1km)² = 3.61 10¹⁵ m²
h = V / A
h = 4.676 10¹⁶ / 3.61 10¹⁵
h = 12.95 m
The correct answer is D
A hammer is thrown from the ground level of the Moon with a speed v0 = 10 m/s at an angle 45◦ above the horizontal. Assume that the acceleration due to gravity on the surface of the Moon is one sixth of the acceleration due to gravity on the surface of the Earth (take g = 10 m/s2 ).(i) Draw a coordinate system and sketch the hammer’s trajectory.(ii) Find the maximum height of flight of the hammer.(iii) Find the component of the hammer’s acceleration vector that is perpendicular to the hammer’s velocity vector at the highest point.(iv) What is the hammer’s velocity at the highest point?
Answer:
Explanation:
initial velocity, vo = 10 m/s
angle of projection, θ = 45°
acceleration due to gravity, g' = g/ 6 = 1.67 m/s^2
(i) diagram is attached.
(ii) The formula for the maximum height is given by
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g'}[/tex]
[tex]H=\frac{10^{2}Sin^{2}45 }{2\times 1.67}[/tex]
H = 14.97 m
(iii) The acceleration of hammer is verticall downwards which is equal to g' = 1.67 m/s^2.
(iv) At heighest point, the hammer has horizontal component of velocity.
V = vo Cos 45 = 7.07 m/s
Answer:
Answer:
i. For the diagram of the hammer's trajectory, it is attached as an image to this solution
u = 10, θ=45°, g = 1/6 * 10 = 1.66667
ii. H = maximum height = [tex]H = \frac{u^{2}sin^{2}\theta }{2g}[/tex]
H = 10*10*(sin45)*(sin45)/2*1.66667
H = 14.9999m
iii. [tex]v^{2} = u^{2} + 2gH\\\\[/tex]
but v = 0 at max height
[tex]v^{2} = 2gH\\=> g = \frac{v^{2} }{2H}[/tex]
g = [tex]\frac{10^{2} }{2*14.9999}[/tex]
[tex]g = 3.3334ms^{-2}[/tex]
iv.
The hammer's velocity at highest point is 0.
Explanation:
For the diagramatic illustration of the problem, it is attached as an image file.
The next thing to be done is to extract the available values for calculation. Now the initial velocity of the hammer is zero since it was thrown from the ground level of the moon
The acceleration due to gravity used in calculation was that of the moon which according to the question is 1/6 that of the moon.
Calculating the maximum height of the hammer just involved substituting the right values in the formula for a projectile's maximum height
To calculate the hammer's acceleration at the maximum height, we need to remember that the velocity of an object at maximum height is always zero. This is because at this instant, the velocity is reduced by the acceleration due to gravity before direction of the object changes. Now, we make use of the formula: [tex]v^{2} = u^{2} + 2gH[/tex]. Now, we substitute v=0 because at maximum height, the velocity is 0. Our aim now is to find g. we substitute the values and remember we calculated the value of the maximum height earlier whose value we will be using in the calculation of g
The answer is zero. This is because at this instant, the velocity is reduced by the acceleration due to gravity before direction of the object changes.
A large crate filled with physics laboratory equipment must be moved up an incline onto a truck The crate is at rest on the incline. What can you say about the force of friction acting on the crate?A. the frictional force points up the inclineB. the frictional force points down the inclineC. the frictional force is zero
Answer:
A. The frictional force points up the incline.
Explanation:
If the crate is at rest, this means that no net force is acting on it.
In absence of friction, we have two forces acting on the crate; the normal force (which prevents that the crate fall through the incline surface), which is always perpendicular to the surface and upward) and gravity force (always downward).
If no net force acting, the 2nd Newton's Law can be expressed as follows:
Fnet = m*a = 0
This is a vector equation, which can be expressed as two algebraic equations, just decomposing both forces along two axes ,perpendicular each other.
As normal force (Fn) has no component along the incline, we can choose as our axes, one parallel to the incline, and another perpendicular, as we will only need to decompose Fg in two perpendicular components.
If the net force is 0, both components must be zero also:
Fx (calling x-axis to the parallel to the incline) = 0
Along the incline. we have a component of the gravity force only, as follows:
Fx = m*g*sin θ = 0
In the perpendicular direction, we have:
Fy = N-m*g*cosθ = 0 ⇒ N = m*g*cos θ (choosing the upward direction as positive)
Returning to Fx, it is clear, that for any angle θ other than 0º, there must be another force, that added to this force, gives 0 as result.
This force can't be another than friction force, which always opposes to the relative movement between the surfaces in contact each other.
As the component of gravity force along the incline tries to accelerate the crate downwards the incline, friction force must point up the incline, i.e., A is the right choice.
When moving a crate up an incline, the frictional force points down the incline to oppose the motion.
The frictional force resists motion, so when the crate is being moved up an incline, the frictional force points down the incline to oppose the motion. Therefore, option B: the frictional force points down the incline.
A welder using a tank of volume 7.60×10−2 m3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.05×105 Pa and temperature of 35.4 ∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 20.8 ∘C, the gauge pressure of the oxygen in the tank is 1.95×105 Pa.Find the initial mass of oxygen. (in Kg)
Answer:
The initial mass of oxygen is 0.38538 Kg.
Explanation:
Given that,
Volume of tank [tex]V= 7.60\times10^{-2}\ m^3[/tex]
Molar mass = 32.0 g/mol
Gauge pressure [tex]P_{g}=3.05\times10^{5}\ pa[/tex]
Initial temperature = 35.4°C
Final temperature = 20.8°C
Gauge pressure [tex]P_{g}=1.95\times10^{5}\ Pa[/tex]
We need to calculate initial pressure
Using formula of pressure
[tex]P_{1}=P_{atm}+P_{g}[/tex]
Put the value into the formula
[tex]P_{1}=1.013\times10^{5}+3.05\times10^{5}[/tex]
[tex]P_{1}=4.063\times10^{5}\ Pa[/tex]
We need to calculate the number of moles
Using equation of ideal gas
[tex]PV= nRT[/tex]
[tex]n_{i}=\dfrac{P_{1}V_{1}}{RT_{1}}[/tex]
Put the value into the formula
[tex]n_{i}=\dfrac{4.063\times10^{5}\times7.60\times10^{-2}}{8.314\times(273+35.4)}[/tex]
[tex]n_{i}=12.043\ moles[/tex]
We need to calculate the initial mass of oxygen
Using formula of mass
[tex]M=m_{m}n_{i}[/tex]
Put the value into the formula
[tex]M=32.0\times12.043[/tex]
[tex]M=385.38\ g[/tex]
[tex]M=0.38538\ kg[/tex]
Hence, The initial mass of oxygen is 0.38538 Kg.
Final answer:
The initial mass of oxygen in the tank can be calculated using the ideal gas law under the given conditions. After calculating the number of moles of oxygen, it is found to be 9.00 moles, which corresponds to an initial mass of 288 g or 0.288 kg.
Explanation:
To find the initial mass of oxygen in the tank, we can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
The initial conditions given are a volume (V) of 7.60\u00d710⁻² m³, a gauge pressure (P) of 3.05\u00d7105 Pa, and a temperature (T) of 35.4 \u00b0C which is 308.55 K (converted from Celsius to Kelvin by adding 273.15).
The ideal gas constant (R) in SI units is 8.314 J/(mol\u00b7K).
Let's perform the calculations:
n = (3.05\u00d710⁵ × 7.60\u00d710⁻²) / (8.314 × 308.55) = 9.00 moles (rounded to three significant figures)
Initial mass = 9.00 moles × 32.0 g/mol = 288 g
Initial mass in kg = 288 g / 1000 = 0.288 kg
Therefore, the initial mass of oxygen in the tank was 0.288 kg.
A wooden plaque is in the shape of an ellipse with height 30 centimeters and width 22 centimeters. Find an equation for the ellipse and use it to find the horizontal width, in centimeters, of the plaque at a distance of 4 centimeters above the center point.
The equation for the given ellipse is (x^2/11^2) + (y^2/15^2) = 1. To find the horizontal width of the plaque at a distance of 4 centimeters above the center point, substitute y = 4 into the equation and solve for x.
Explanation:To find an equation for the ellipse, we can use the standard form equation: (x^2/a^2) + (y^2/b^2) = 1, where a is the semi-major axis and b is the semi-minor axis. In this case, the height of the ellipse is the semi-major axis and the width is the semi-minor axis. So, the equation for the ellipse is: (x^2/11^2) + (y^2/15^2) = 1.
To find the horizontal width of the plaque at a distance of 4 centimeters above the center point, we can substitute the y-coordinate as 4 into the equation and solve for x. The equation becomes: (x^2/11^2) + (4^2/15^2) = 1. Solving this equation will give us the horizontal width (x) of the plaque at that distance.
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what?
Answer:
v₂ = 176.24 m/s
Explanation:
given,
angle of projectile = 45°
speed = v₁ = 150 m/s
for second trail
speed = v₂ = ?
angle of projectile = 37°
maximum height attained formula,
[tex]H_{max}= \dfrac{v^2 sin^2(\theta)}{g}[/tex]
now,
[tex]H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}[/tex]
[tex]H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}[/tex]
now, equating both the equations
[tex]\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}[/tex]
[tex]\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}[/tex]
v₂² = 31061.79
v₂ = 176.24 m/s
velocity of projectile would be equal to v₂ = 176.24 m/s
When a sound wave travels directly toward a hard wall, the incoming and reflected waves can combine to produce a standing wave. There is an antinode right at the wall, just as at the end of a closed tube, so the sound near the wall is loud. You are standing beside a brick wall listening to a 85 Hz tone from a distant loudspeaker.How far from the wall must you move to find the first quiet spot?
The student should position themselves approximately 1.009 meters from the brick wall to find the first quiet spot, which is the location of the first node in the standing wave pattern created by the 85 Hz sound wave reflecting off the wall.
When a sound wave travels directly toward a hard wall, it reflects and can combine with the incoming wave to create a standing wave pattern. At the position of a node in this pattern, there is destructive interference, and the sound is minimized or goes silent. Since the wall acts like the closed end of a tube, we know there's an antinode at the wall, and the distance from the wall to the first node would be one quarter of the wavelength of the sound.
The wavelength (λ) can be found using the speed of sound in air (approximately 343 meters per second) and the frequency of the tone. The equation v = f λ gives us λ = v / f. Plugging in the values, we get λ = 343 m/s / 85 Hz = 4.035 meters. Thus, the distance from the wall to the first quiet spot, which is one quarter of the wavelength, is λ / 4 = 4.035 m / 4 ≈ 1.009 meters.
The student should move approximately 1.009 meters away from the wall to find the first quiet spot where there is a node of a standing wave created by the 85 Hz tone.
A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is, V = ax + b, where a and b are constants. If the flow is constant with V1 = 10 m/s at x1 = 0 and V2 = 25 m/s at x2 = 1 m, determine (a) the local acceleration; (b) the acceleration magnitude of the fluid at point (1); (c) the acceleration magnitude of the fluid at point (2).
Answer:
Explanation:
v = ax + b
when x = 0 , v = 10 ( given )
10 = b
when x = 1 , v = 25
25 = a + b
25 = a + 10
a = 15
So
v = 15 x + 10
acceleration
v = 15 x + 10
differentiating it on both sides wrt t
a = dv / dt = 15 dx / dt
a = 15 v = 15 ( 15x + 10 )
a = 225x + 150
acceleration at x = 0
a = 150 m/s²
acceleration at x = 1 m
a = 225 + 150
= 375 m /s²
The value of a is 15. The acceleration magnitude at point (1) is equal to the value of a, which is 15 m/s². The acceleration magnitude at point (2) is also equal to the value of a, which is 15 m/s².
Given:
V₁ = 10 m/s
V₂ = 25 m/s
(a) To find the value of V = ax + b, we can use the velocity values at x₁ and x₂:
V1 = a(0) + b (since x₁ = 0)
V2 = a(1) + b (since x₂ = 1)
10 = b
25 = a + b
25 = a + 10
a = 15
Therefore, the value of a is 15.
(b) To find the acceleration magnitude at point (1), we differentiate the velocity equation with respect to x:
a = dv/dx
a = d(ax + b)/dx
a = a
So, the acceleration magnitude at point (1) is equal to the value of a, which is 15 m/s².
(c) To find the acceleration magnitude at point (2):
a = d(ax + b)/dx
a = a
Therefore, the acceleration magnitude at point (2) is also equal to the value of a, which is 15 m/s².
To know more about acceleration:
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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.
Answer:
-4.40
Explanation:
explanation is in attachment
To determine Fx, the x-component of the force, we consider the torque formula τ = r × F, solve for Fx using the given torque and position vectors, and by calculating the determinant involving the unit vectors and the components of r and F.
Explanation:The student is asked to calculate the value of Fx from a given torque value and the position vector of the particle. In physics, particularly in the study of dynamics, the torque about a point is the cross product of the position vector (r) and the force (F) applied. The torque (τ) vector formula is:
τ = r × F
Given the torque vector τ = (3.40 N · m)i + (2.80 N · m)j + (0.800 N · m)k, and the position vector r = (2.00 m)i - (3.00 m)j + (2.00 m)k, we can set up the equation to solve for the torque:
τ = r × F ==> (3.40 N · m)i + (2.80 N · m)j + (0.800 N · m)k = (2.00 m)i - (3.00 m)j + (2.00 m)k × (Fxi + 7.00 Nj - 5.80 Nk)
We can calculate it using the determinant of a 3x3 matrix consisting of the unit vectors i, j, and k, the components of r, and the components of F. Solving for Fx involves calculating the determinant and equating each corresponding component of the torque vector to form a system of equations. The x-component of the torque provides the equation to solve for Fx, as the other components give us irrelevant information for this specific calculation.