Answer:
T = 9.19 seconds
Explanation:
It is given that,
Length of the pendulum, l = 21 m
We need to find its time period. We know that the time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{L}{g}}[/tex]
g is acceleration due to gravity
So,
[tex]T=2\pi \sqrt{\dfrac{21}{9.8}}\\\\T=9.19\ s[/tex]
So, the period of the pendulum is 9.19 seconds.
Final answer:
The period of the Smithsonian's 21.0 m long pendulum is approximately 9.2 seconds, which is calculated using the formula for a simple pendulum's period T = 2π√(L/g).
Explanation:
The period of a pendulum primarily depends on its length and the acceleration due to gravity. Using the formula for the period of a simple pendulum T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), we can calculate the period of the Smithsonian's pendulum.
Given that the length L is 21.0 m and by taking the standard value for g, 9.81 m/s², we plug these values into the formula to find the period:
T = 2π√(21.0 m / 9.81 m/s²)
Calculating this we get:
T ≈ 2π√(2.14 s²)
T ≈ 2π√(2.14 s²)
T ≈ 2π(1.46 s)
T ≈ 9.2 s
This means the Smithsonian's pendulum would have taken approximately 9.2 seconds to complete one full swing, back and forth. This is the time it would take to go from one side to the other and back again, corresponding to one complete oscillation.
For an experiment, a student wants to maximize the power output for a circuit. The student has three resistors of resistances 512, 1012, and 20 S, and a 5 V battery. Which of the following best describes how the students should arrange the resistors so that when they are connected to the battery it will maximize the power output for the circuit and explains why? a) The student should place all resistors in parallel, because the battery supplies more current to a parallel combination than to a series combination b) The student should place all resistors in parallel, because the potential difference across all the resistors is less for a parallel combination than for a series combination c) The student should place all resistors in series, because this creates the largest equivalent resistance d) The student should place all resistors in series, because the battery supplies more current to a series combination than to a parallel combination e) The student should place all resistors in series, because the potential difference across all the resistors is greater in series than in parallel
Final answer:
To maximize the power output of a circuit with a 5V battery and three resistors, the resistors should be placed in parallel because this arrangement supplies more current than a series combination, thereby maximizing power.
Explanation:
To maximize the power output for a circuit with a given voltage source, the resistors should be arranged to draw the highest current possible. For resistors connected in parallel, the total resistance of the circuit decreases, as opposed to when they are in series, where it increases. Since power (P) is calculated using the formula P = V^2 / R, where V is the voltage and R is the resistance, minimizing R will maximize P.
Because a parallel configuration guarantees that each resistor gets the full voltage of the battery, the correct answer is: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination. This is because in a parallel circuit, the voltage across each resistor is equal to the voltage of the source, which maximizes the current through each resistor due to Ohm's law (I = V/R), and therefore maximizes the total current in the circuit.
Answer option (a) is thus correct: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination.
A 0.5 Kg pinball is initially at rest against a 120 N/m spring. The shooter is pulled back and has the spring compressed a distance of 0.2 m. The spring is released and the ball is shot up the ramp. It hits nothing and eventually comes to rest before it begins to roll down. We can ignore friction. The game board ramp is at an angle of 30o . How far did the ball travel on the board from the place of maximal compression to the first stop
Answer:
[tex]\Delta s = 0.978\,m[/tex]
Explanation:
The pinball-spring system is modelled after the Principle of Energy Conservation:
[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]
[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]
The height reached by the pinball is:
[tex]\Delta h = 0.489\,m[/tex]
The distance travelled by the pinball is:
[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]
[tex]\Delta s = 0.978\,m[/tex]
wo plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.70 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 500-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
What is the magnitude of the dipole moment of the arrangement?
Answer:
the magnitude of the dipole moment is 3.5*10^-11Cm
Explanation:
The dipole moment is given by the following formula:
[tex]\mu=qr[/tex]
r: distance between the centers of the charges = 500mm
q: charges of the bowling balls = 0.70nC
By replacing you obtain:
[tex]\mu=(0.70*10^{-9}C)(500*10^{-4}m)=3.5*10^{-11}Cm[/tex]
hence, the magnitude of the dipole moment is 3.5*10^-11Cm
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.50 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.500 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.52 m and t = 0.150 s .
Answer:
0.0549 m
Explanation:
Given that
equation y(x,t)=Acos(kx−ωt)
speed v = 8.5 m/s
amplitude A = 5.5*10^−2 m
wavelength λ = 0.5 m
transverse displacement = ?
v = angular frequency / wave number
and
wave number = 2π/ λ
wave number = 2 * 3.142 / 0.5
wave number = 12.568
angular frequency = v k
angular frequency = 8.5 * 12.568
angular frequency = 106.828 rad/sec ~= 107 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)= 5.5*10^−2 cos(12.568 x−107t)
when x =0 and and t = 0
maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))
maximum y(x,t)= 5.5*10^−2 m
and when x = x = 1.52 m and t = 0.150 s
y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )
y(x,t)= 5.5*10^−2 × (0.9986)
y(x,t) = 0.0549 m
so the transverse displacement is 0.0549 m
When you hear a noise, you usually know the direction from which it came even if you cannot see the source. This ability is partly because you have hearing in two ears. Imagine a noise from a source that is directly to your right. The sound reaches your right ear before it reaches your left ear. Your brain interprets this extra travel time (Δt) to your left ear and identifies the source as being directly to your right. In this simple model, the extra travel time is maximal for a source located directly to your right or left (Δt = Δtmax). A source directly behind or in front of you has equal travel time to each ear, so Δt = 0. Sources at other locations have intermediate extra travel times (0 ≤ Δt ≤ Δtmax). Assume a source is directly to your right.(a) Estimate the distance between a person's ears. (they gave us the answer of .2... apparently the program is messed up and we have to use .2(b) If the speed of sound in air at room temperature is vs = 338 m/s,find Δtmax. (Use your estimate.)
(c) Find Δtmax if instead you and the source are in seawater at the same temperature, where vs = 1534 m/s. (Use your estimate.)
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The average distance calculated between a person's ears is 0.2 meters. The interaural time difference (ITD) is maximal when the sound source is directly to the side. For air (vs=338 m/s), Δ[tex]t_{max}[/tex] is 591.72 microseconds, and for seawater (vs= 1534 m/s), it is 130.38 microseconds. These calculations illustrate sound localization and the impact of medium on sound propagation.
Understanding Sound Localization and Interaural Time Difference
When a noise occurs directly to the right of a person, the sound waves reach the right ear before the left ear. We can calculate the maximum interaural time difference ( Δ[tex]t_{max}[/tex]) for sound reaching the ears using the given distance between the ears.
(a) The average distance between a person's ears was estimated as 0.2 meters (20 cm).
(b) To calculate Δ[tex]t_{max}[/tex] with the speed of sound in air (vs = 338 m/s), we can use the formula Δ[tex]t_{max}[/tex] = d / vs, where d is the distance between ears. Substituting the values, we get:
Δ[tex]t_{max}[/tex] = 0.2 m / 338 m/s = 0.0005917159763 seconds, or approximately 591.72 microseconds.
(c) Lastly, for sound traveling in seawater at room temperature where vs = 1534 m/s, we similarly get:
Δ[tex]t_{max}[/tex] = 0.2 m / 1534 m/s = 0.0001303763441 seconds, or approximately 130.38 microseconds.
This demonstrates the role of the medium in sound propagation and how it affects the interaural time difference.
When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at speed V.
If the rocket doubles its speed to 200 m/s, the sound waves from the engine will now approach the mountain at speed
A. 2–√V
B. 2V
C. V
D. 4V
Answer:
The correct option is C
Explanation:
From the question we are told that
The initial speed of the rocket is [tex]v_i = 100 m/s[/tex]
The speed of the rocket engine sound is [tex]V[/tex]
The final speed of the rocket is [tex]v_f = 200 \ m/s[/tex]
The speed of the sound at [tex]v_f[/tex] would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).
A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning
Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates
Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of the ratio of the electrical force and gravitational force between an electron and a proton separated by a distance of 1 m.
1. What is the magnitude of the electrical force?
The Coulomb constant is 8.98755 x 10^9 N*m^2/C^2 , the gravitational constant is 6.67259 x 10^11 m^3 /kg*s^2 , the mass of a proton is 1.67262 x 10^-27 kg, the mass of an electron is 9.10939 x 10^−31 kg, and the elemental charge is 1.602 x 10^-19 C. Answer in units of N.
Explanation:
The electrical force between charges is given by :
[tex]F_e=\dfrac{kq_eq_p}{r^2}[/tex]
[tex]q_e\ and\ q_p[/tex] are charge on electron and proton respectively.
[tex]F_e=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\\F_e=2.3\times 10^{-28}\ N[/tex]
The Gravitational force between masses is given by :
[tex]F_G=\dfrac{Gm_em_p}{r^2}[/tex]
[tex]m_e\ and\ m_p[/tex] are masses of electron and proton respectively.
[tex]F_G=\dfrac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}{1^2}\\\\F_G=1.01\times 10^{-67}[/tex]
Ratio of electrical to the gravitational force is :
[tex]\dfrac{F_e}{F_G}=\dfrac{2.3\times 10^{-28}\ N}{1.01\times 10^{-67}\ N}\\\\\dfrac{F_e}{F_G}=2.27\times 10^{39}[/tex]
Hence, this is the required solution.
Final answer:
To find the magnitude of the electrical force between an electron and proton separated by 1 m, we use Coulomb's Law with given constants. The electrical force is calculated to be approximately 2.30 x 10⁻¹⁰ N.
Explanation:
To calculate the magnitude of the electrical force between an electron and proton separated by 1 meter, we use Coulomb's Law which is given by the formula:
F = k * |q¹ * q²| / r²
Where F is the force in Newtons (N), k is the Coulomb constant (8.98755 x 10⁹ N*m²/C²), q1 and q2 are the charges of the proton and electron respectively, and r is the separation distance in meters. Since both the proton and the electron have an elemental charge of 1.60² x 10⁻¹⁹ C, albeit with opposite signs, their charges can be multiplied to give their product in Coulomb's equation. The separation r is 1 meter.
So the magnitude of the electrical force F is calculated as:
F = (8.98755 x 109 N*m²/C²) * (1.60² x 10⁻¹⁹ C)² / 12m²
F ≈ 2.30 x 10¹⁰ N
Use the drop-down menus to complete the scenarios.
A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing
treatment for it. There may be other tumors and a potential blockage in the surrounding area that need to be
investigated. The imaging technique that might provide the most information in this case is
Joe has ongoing issues with his throat and feels some sort of blockage or abnormality as he swallows. The doctor
decides to use X-ray imaging to visualize Joe's internal anatomy as he swallows to help determine the nature of the
problem.
will be used for this procedure.
First: CT Scan
Second: Fluoroscopy
Explanation:
Both correct
A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.
Explanation:In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.
In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.
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Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop. There is no friction between block A and tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string. What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest
The largest mass that block C can have so that blocks A and B still slide together, given the values, is approximately 10.2 kg.
Explanation:The subject of this question is a physics problem involving static friction, mass, and gravity. The largest mass that block C can have so that blocks A and B still slide together can be calculated using the principle of static friction and the relevant equation:
fs_max = µs (mA + mB)g, with fs_max representing the maximum force of static friction, µs=0.75 being the coefficient of static friction, and g=9.8 m/s² being the acceleration due to gravity. Here mA=8 kg and mB=5 kg are the masses of block A and B respectively.
To keep A and B together, the tension (T) in the string must be less than or equal to fs_max. As T is also equal to the weight (mCg) of block C, from where we can find mC ≤ fs_max / g. Substituting the given values and calculations, we find that the maximum mass of block C should be around 10.2 kg for blocks A and B to slide together.
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Final answer:
The maximum mass that block C can have to prevent sliding between block A and B is calculated using the static friction coefficient [tex](\mu_s)[/tex] and the weight of block B [tex](m_B)[/tex], which results in a mass of 3.75 kg for block C.
Explanation:
To figure out the largest mass that block C can have without block A and B sliding relative to each other, we must use the static friction force and Newton's second law. The static friction force [tex](f_s)[/tex] is what keeps block B from sliding on block A.
This force is given by [tex]f_s = \mu_s \times N[/tex], where μ_s is the coefficient of static friction and N is the normal force. Since block B is at rest on block A, N is equal to the weight of block B, [tex]N = m_B \times g[/tex].
For block A and block B to accelerate together, the tension in the string (T) caused by the weight of block C must not exceed the static friction force. Therefore, the maximum force that can be applied by block C before block B starts sliding is the static friction force: T ≤ f_s.
So, the maximum weight (and hence mass) that block C can have is when [tex]T = f_s[/tex]. Since [tex]T = m_C \times g[/tex] for block C, we have m_C × g = μ_s × m_B × g. Canceling out g and solving for m_C gives us the formula [tex]m_C = \mu_s \times m_B[/tex].
Substituting the given values, we get [tex]m_C = 0.750 \times 5.00 kg = 3.75 kg[/tex]. Hence, the largest mass that block C can have without causing sliding between block A and B is 3.75 kg.
The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open square brackets fraction numerator x over denominator 0.35 space straight m end fraction close square brackets. What is the angular frequency of small oscillations around the point x = 0?
Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
Characteristics help scientists ________ objects.
Answer:Identify
Explanation:
Answer:
b
Explanation:
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver
Determining the radius of the satellite TV dish to achieve a specific electric field requires knowing the intensity and power of the broadcast signal as well as the dish's area. The radius can be calculated from the formula for the area encompassed by the dish, but without additional information about the broadcast power or spread area, a specific radius cannot be provided.
Explanation:To determine how large the radius R of the satellite TV receiver dish must be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver, we need to relate the intensity of the electromagnetic wave to the electric field amplitude and the area of the dish. The intensity (I) is related to the electric field strength (E) by the relationship I = rac{1}{2} extZ_0[tex]E^2[/tex], where Z0 is the impedance of free space (approximately 377 ohms).
The power received by the dish (Pr) is the product of the intensity and the area of the dish, Ad: Pr = I imes Ad. Given that the receiver has an area of 5 cm2 and the required electric field amplitude is 0.1 mV/m, we would solve for the radius R of the dish using the formula for area of a circle, A = \\(pi)[tex]R^2[/tex].
However, to solve this problem, we would need additional information such as the power broadcast by the satellite and over what area this power is spread. With our current information, we cannot provide an exact answer, but typically residential satellite dishes have diameters a little less than half a meter to effectively receive TV signals.
n optician is performing Young's double-slit experiment for her clients. She directs a beam of monochromatic light to a pair of parallel slits, which are separated by 0.134 mm from each other. The portion of this light that passes through the slits goes on to form an interference pattern upon a screen, which is 4.50 meters distant. The light is characterized by a wavelength of 553 nm. (a) What is the optical path-length difference (in µm) that corresponds to the fourth-order bright fringe on the screen? (This is the fourth fringe, not counting the central bright band, that one encounters moving from the center out to one side.)
Answer:
Explanation:
This is a problem based on interference pf light waves.
wavelength of light λ = 553 nm
slit separation d = .134 x 10⁻³ m
screen distance D = 4.5 m
for fourth order bright fringe, path- length difference = 4 x λ
= 4 x 553
= 2212 nm .
= 2.212 μm
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.1 cm ( 0.581 m) and the flow speed of the petroleum is 10.1 m/s. At the refinery, the petroleum flows at 5.85 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?
volume flow rate:________m^3/sdiameter________cm
Answer:
volume flow rate:2.68m³/s
diameter :58.27cm
Explanation:
flow rate = Q = πr² v = amount per second that flows through the pipe
Given:
pipe's diameter= 0.581
r= 0.581/2=>0.2905m
speed 'v'= 10.1 m/s
Q= (3.142)(0.2905)²(10.1)
so
volume flow rate= 2.68m³/s
->if no oil has been added or subtracted or compressed then Q is the same everywhere
therefore,
Q= πr²v
2.68 = π(d/2)²(5.85)
d= (2.68x4) /(3.142 x 5.85)
d= 0.5827m =>58.27cm
A diffraction grating is illuminated with yellow light. The diffraction pattern seen on a viewing screen consists of three yellow bright spots, one at the central maximum (θ = 0°) and one on either side of it at θ = ±50°. Then the grating is simultaneously illuminated with red light. Where a red and a yellow spot overlap, an orange spot is produced. The new pattern consists of __________
Answer:
an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.
Explanation:
In the visible spectrum- red to violet, red has the highest wavelength.
The maximum internsity for diffraction grating is given by,
Sinθ[tex](_{max} )[/tex] = mλ/d
It is concluded that the angle of diffraction increases with increase in wavelength'λ' . So, red fringe will be farthest from the center, orange light will be at the center and yellow fringe will be at 50°.
Therefore, The new pattern consists of : an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.
When you run around a track at 5 km/h, your velocity is constant. Please select the best answer from the choices provided T F
Answer:
T
Explanation:
Answer:
true
Explanation:
Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjacent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance a=33.7 cm from the wall. The reflected beam strikes the wall at a height b=36.7 cm above the surface of the mirror. Find the angle of incidence θi at which the laser beam strikes the mirror.
Answer:
The angle of incidence is [tex]\theta_i =42.6^o[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The distance between the mirror and the wall is [tex]a = 33.7 cm[/tex]
The height of the above the mirror is [tex]b = 36.7 cm[/tex]
Generally the angle which the reflected ray make with the mirror is mathematically evaluated as
[tex]\alpha =tan ^{-1} (\frac{b}{a})[/tex]
substituting values
[tex]\alpha = tan ^{-1}( \frac{36.7}{33.7})[/tex]
[tex]\alpha =47.4^o[/tex]
From the diagram we can deduce that the angle of incidence is
[tex]\theta_i = 90 - \alpha[/tex]
So [tex]\theta_i = 90 - 47.4[/tex]
[tex]\theta_i =42.6^o[/tex]
A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s
to the west. The launching rod strikes the pinball and causes it to move in the
opposite direction with a velocity of 10.0 m/s. What impulse was delivered to the
pinball by the launcher?
A 0.75 kg•m/s to the east
B 1.2 kg•m/s to the east
C 1.8 kg•m/s to the east
D 3.0 kg•m/s to the east
The impulse delivered to the pinball by the launcher is 1.8 kg•m/s to the east
Impulse:Let us consider east as positive direction and west as negative direction.
Then, from the question, we get that the initial velocity of the pinball is
2m/s towards the west
or u = - 2 m/s
and the mass of the pinball is m = 150g = 0.15 kg
So, the initial momentum of the pinball is:
[tex]P_i=mu\\\\P_i=0.15\times(-2)\;kgm/s\\\\P_i=-0.3\;kgm/s[/tex]
Now, the final velocity of the pinball after being struck by the rod is 10 m/s towards the east,
or v = 10 m/s
So, the final momentum of the pinball is:
[tex]P_f=mu\\\\P_f=0.15\times(10)\;kgm/s\\\\P_f=1.5\;kgm/s[/tex]
Impulse is defined as the change in momentum, that is,
[tex]I=\Delta P\\\\I=P_f-P_i\\\\I=1.5-(-0.3)\\\\I=1.8\;kgm/s[/tex]
The impulse is 1.8 kgm/s towards the east.
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A cart of mass 300 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.0 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. Correct: Your answer is correct. s (b) Find its maximum speed. Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position. m/s
Answer:
Explanation:
the angular frequency ω of the pendulum is given by the formula
ω = [tex]\sqrt{\frac{k}{m} }[/tex] , k is spring constant , m is mass attached .
= [tex]\sqrt{\frac{9}{.3} }[/tex]
= 5.48 rad /s
time period = 2π / ω
= 2 x 3.14 / 5.48
= 1.146 s
b ) formula for speed
v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex] , a is amplitude , x is displacement from equilibrium point.
for maximum speed x = 0
max speed = ωa
= 5.48 x 3.8 x 10⁻² ( initial displacement becomes amplitude that is 3.8 cm )
= .208 m /s
20.8 cm / s
c )
when x = .02 m , velocity = ?
v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]
= 5.48 [tex]\sqrt{(.038^2-\ .02^2)}[/tex]
= 5.48 x .0323109
= .177 m /s
17.7 cm /s .
To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length of the tube, as illustrated below. In this time she heats the tube electrically, passing a current of 50 mA through the tube wall. The initial temperature of the tube is 25 C. After 10 min of heating, a thermocouple buried inside the tube wall measures a temperature of 31.2 C. The tube core is sealed and the heat lost from the internal surface of the tube is negligible during measurements. Furthermore, the tube is suspended on two thermally insulating supports. What is the specific heat of the tube?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Learning Goal: To understand how to find the wavelength and diffraction patterns of electrons. An electron beam is incident on a single slit of width aaa. The electron beam was generated using a potential difference of magnitude VVV. After passing through the slit, the diffracted electrons are collected on a screen that is a distance LLL away from the slit. Assume that VVV is small enough so that the electrons are nonrelativistic. Ultimately, you will find the width of the central maximum for the diffraction pattern.
Final answer:
The question involves calculating the wavelength and diffraction patterns of electrons in a single slit experiment. The student would use the de Broglie wavelength formula and the principles of single slit diffraction to find the dimensions of the central maximum. It demonstrates the wave-particle duality exhibited by electrons.
Explanation:
The student is seeking to understand how to find the wavelength and diffraction patterns of electrons. The context involves an electron beam passing through a single slit, creating a diffraction pattern on a distant screen. From the given potential difference, one could calculate the de Broglie wavelength of the electrons, as their velocity can be determined under the assumption that they are nonrelativistic. The central maximum's width on the diffraction pattern can be deduced using the slit width and the wavelength.
The phenomenon of diffraction and interference highlighted in the question is a demonstration of the wave-like properties of electrons, referred to as wave-particle duality. The experimental setup often includes narrow slits whose sizes are comparable to the wavelength of electrons, resulting in observable wave effects such as constructive and destructive interference.
The angular positions of the minima and maxima in the diffraction pattern are crucial for determining the dimensions of the pattern. The de Broglie wavelength plays a significant role in these calculations, linking the microscopic quantum world to observable macroscopic patterns.
Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-Hz sound waves into a room where the speed of sound is 340 m/s. Consider a point 4.0 m in front of speaker 1, which lies along a line from speaker 1, that is perpendicular to a line between the two speakers. Is this a point of maximum constructive interference, a point of perfect destructive interference, or something in between?
Answer:
It is somewhere in between
Explanation:
Wave length of sound from each of the speakers = 340 / 1700 = 0.2 m = 20 cm
Distance between first speaker and the given point = 4 m.
Distance between second speaker and the given sound
D = √(4² + 2²)
D = √(16 + 4)
D = √20
D = 4.472 m
Path difference = 4.472 - 4 =
0.4722 m.
Path difference / wave length = 0.4772 / 0.2 = 2.386
This is a fractional integer which is neither an odd nor an even multiple of half wavelength. Hence this point is of neither a perfect constructive nor a perfect destructive interference.
A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.50 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction
Explanation:
From maxwell's equations;
B = E/v
Where;
B is maximum magnitude of magnetic field
E is maximum electric field
v is speed of light which has a constant value of 3 x 10^(8) m/s
We are given, E = 4.5 V/m
Thus; B = 4.5/(3 x 10^(8))
B = 1.5 x 10^(-8) T
Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction
Thus,
Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction
Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex] in the positive y-direction.
From Maxwell's equations,
[tex]\bold {B = \dfrac Ev}[/tex]
Where;
B - maximum magnitude of magnetic field = ?
E- maximum electric field = 4.5 V/m
v- speed of light = 3 x 10^(8) m/s
Put the values in the formula,
[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]
When Electric field, is in the positive x-direction, vector B is in the positive y-direction and the product of vector E and vector B will have to be in the positive z-direction.
Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex] in the positive y-direction.
To know more about Maxwell's equation,
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The air in a car tire la compressed when the car rolls over a rock. If the air
outside the tire becomes slightly warmer, what are the correct signs of heat
and work for this change?
Answer:
the signs of heat and work are; -Q and -W
Explanation:
The first law of thermodynamics is given by; ΔU = Q − W
where;
ΔU is the change in internal energy of a system,
Q is the net heat transfer (the sum of all heat transfer into and out of the system)
W is the net work done (the sum of all work done on or by the system).
Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).
Since work is done by the system, W remains negative.
Thus, the signs of heat and work are; -Q and - W
Describe the process of nuclear fission. How does it work and when is it used? What is meant by a chain reaction, and how do you control it?
Does this chemical equation support the Law of Conservation of Mass? Why or why not?
N₂ + H₂ → NH₃
Answer:No it does not the chemical equation is unbalanced,this is because according to the law of conservation it states that mass cannot be created or destroyed ,the chemical equation N₂ + H₂ → NH₃ shows that energy has been created which is not possibly so the correct balanced chemical is equation is 1N₂ +3H₂ → 3NH₃
Explanation:
A stone is moving on a rough level surface. It has 24 J of kinetic energy, and the friction force on it is a constant 0.50 N. What is the maximum distance it can slide?
The energy an object has as a result of motion is known as kinetic energy.A force must be applied to an object in order to accelerate it.We must put in effort in order to apply a force.After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.
Solve the problem ?
We have provided This is supposed to be a stone, and it is travelling in this way, on a rough, level surface.It has 24 Joel of initial kinetic energy.And 0.50 Newton is the major force of friction on it.The friction force is this.This equates to 0.50fresh banknotes.What is the longest it can slide before coming to a stop?Assume that it has a sliding distance of D before halting.Therefore, the final candidate energy will be 0.We will now use the idea of energy heroes, of which we are aware. This friction force must be equivalent to the change in candidate energy.Finally, without K. Initiative.As a result, friction force will multiply the work done by frictionAs a result, friction force will multiply the work done by friction.The displacement and force are at an angle of 1 80 degrees in this position.As a result, this has lost him 1 80 since his negative 24.Energy nil because of final counting.Therefore, the cost of this distance D. will be 24 divided by.0.50.Consequently, this is a 48-meter cost alternative.To learn more about kinetic energy refer
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Final answer:
To determine the maximum distance the stone can slide on a rough surface, divide the kinetic energy by the friction force. The maximum distance is 48 meters.
Explanation:
To determine the maximum distance the stone can slide on a rough surface, we need to calculate the work done by the friction force.
Work done = Force x Distance
Given that the friction force is 0.50 N and the stone has 24 J of kinetic energy, we can rearrange the equation to solve for distance:
Distance = Kinetic energy / Friction force
Plugging in the values, we get:
Distance = 24 J / 0.50 N = 48 meters
Therefore, the maximum distance the stone can slide on the rough surface is 48 meters.
Write a function to model the volume of a rectangular prism if the length is 24cm and the sum of the width and height is 28cm
[tex]w + h = 28[/tex]
[tex]l \times w \times h = v[/tex]
[tex]v = w \times h \times 24[/tex]
[tex]w = 28 - h[/tex]
[tex]h(28 - h) \times 24 = v[/tex]
[tex]24( - {h}^{2} + 28h) = v[/tex]
A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fuel before liftoff is 25.0 g, which includes fuel mass of 12.7 g. The engine fires for a total of 1.90 s. (Assume all the fuel is consumed.) (a) Calculate the average exhaust speed of the engine (in m/s). (b) This engine is situated in a rocket case of mass 63.0 g. What is the magnitude of the final velocity of the rocket (in m/s) if it were to be fired from rest in outer space with the same amount of fuel
Answer:
a) v = 786.93 m/s
b) v = 122.40 m/s
Explanation:
a) To find the average exhaust speed (v) of the engine we can use the following equation:
[tex] F = \frac{v\Delta m}{\Delta t} [/tex]
Where:
F: is the thrust by the engine = 5.26 N
Δm: is the mass of the fuel = 12.7 g
Δt: is the time of the burning of fuel = 1.90 s
[tex]v = \frac{F*\Delta t}{\Delta m} = \frac{5.26 N*1.90 s}{12.7 \cdot 10^{-3} kg} = 786.93 m/s[/tex]
b) To calculate the final velocity of the rocket we need to find the acceleration.
The acceleration (a) can be calculated as follows:
[tex] a = \frac{F}{m} [/tex]
In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:
[tex]m = \frac{(m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = \frac{2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = \frac{2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g[/tex]
Now, the acceleration is:
[tex] a = \frac{5.26 N}{81.65 \cdot 10^{-3} kg} = 64.42 m*s^{-2} [/tex]
Finally, the final velocity of the rocket can be calculated using the following kinematic equation:
[tex]v_{f} = v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s[/tex]
I hope it helps you!
Final answer:
The average exhaust speed of the engine is calculated to be approximately 786.929 m/s, and the magnitude of the final velocity of the rocket if fired in outer space is roughly 94.916 m/s.
Explanation:
To calculate the average exhaust speed (v_ex), we can use the impulse-momentum theorem, which states that impulse is equal to the change in momentum of the system. Impulse is given by the product of the average force exerted by the engine (thrust) and the time interval during which the thrust is applied. If all the fuel is consumed, the change in mass (Δm) is the mass of the fuel.
Impulse = Thrust × Time = 5.26 N × 1.90 s = 9.994 N·s
The momentum change is equal to the mass of the fuel expelled times the average exhaust speed.
Δ(momentum) = Δm × v_ex
Substituting the impulse and solving for v_ex, we get:
v_ex = Impulse / Δm
v_ex = 9.994 N·s / 0.0127 kg = 786.929 m/s
Part B: Final Velocity of the Rocket in Space
The final velocity (V_final) of the rocket in space can be determined using the rocket equation, also known as Tsiolkovsky's rocket equation:
V_final = v_ex × ln(m_initial / m_final)
Where:
m_initial = mass of the rocket with fuel = 25.0 g + 63.0 g = 88.0 g = 0.088 kgm_final = mass of the rocket without fuel = 25.0 g - 12.7 g + 63.0 g = 75.3 g = 0.0753 kgCalculating the final velocity:
V_final = 786.929 m/s × ln(0.088 kg / 0.0753 kg) ≈ 94.916 m/s
Interference in 2D. Coherent red light of wavelength λ = 700 nm is incident on two very narrow slits. The light has the same phase at both slits. (a) What is the angular separation in radians between the central maximum in intensity and an adjacent maximum if the slits are 0.025 mm apart? (b) What would be the separation between maxima in intensity on a screen located 1 m from the slits? (c) What would the angular separation be if the slits were 2.5 mm apart? (d) What would the separation between maxima be on a screen 25 mm from the slits? The answer is relevant to the maximum resolution along your retina that would be useful given the size of your pupil. The spacing between cones on the retina is about 10 µm. (Of course, you would never want to look directly into a laser beam like this....) (e) How would the location of the maxima change if one slit was covered by a thin film with higher index of refraction that shifted the phase of light leaving the slit by π? Would the separation change? 3. Combined two source interference and
Answer:
Explanation:
λ
given λ = 700 nm
a) for first maxima, d*sin(θ) = λ
sin(theta) = λ/d
= 700*10^-9/(0.025*10^-3)
= 0.028
theta θ = sin^-1(0.028)
= 1.60 degrees
b) given R = 1 m,
delta_y = λ*R/d
= 700*10^-9*1/(0.025*10^-3)
= 0.028 m or 2.8 cm
c) for first maxima, d*sin(θ) = λ
sin(θ) = λ/d
= 700*10^-9/(2.5*10^-3)
= 0.00028
theta = sin^-1(0.00028)
= 0.0160 degrees
d) R = 25 mm = 0.025 m
δ_y = λ*R/d
= 700*10^-9*0.025/(2.5*10^-3)
= 7*10^-6 m or 7 micro m
e) No. But the position of maxima and minima will be shifted.