The frequency of the third normal mode of a guitar string with a length of 40.0 cm, mass of 0.5 g when stretched with a tension of 80N is approximately 948.68 Hz.
To calculate the frequency of the third normal mode (
n = 3) for a guitar string, we can use the formula for the frequency of a string fixed at both ends:
f_n = (n/2L) √(T/μ)
where:
f_n is the frequency of the nth mode,
n is the mode number (which is 3 in this case),
L is the length of the string,
μ is the mass per unit length of the string (linear mass density), and
T is the tension in the string.
Given the length of the string L = 40.0 cm = 0.4 m, the mass m = 0.5 g = 0.0005 kg, and the tension T = 80 N, we first need to calculate the linear mass density:
μ = m/L
In this case,
μ = 0.0005 kg / 0.4 m = 0.00125 kg/m
Now, we use the frequency formula to find f_3:
f_3 = (3/2 0.4 m √(80 N/0.00125 kg/m)
= (3/(0.8 m)) √(64000 N/m)
= 3.75 √(64000 N/m)
= 3.75 √(64000 N/m)
= 3.75 * 252.9822 Hz
= 948.68 Hz
The frequency of the third normal mode of the guitar string is approximately 948.68 Hz.
The theory of nuclear astrophysics is that all the heavy elements like uranium are formed in the interior of massive stars. These stars eventually explode, releasing these elements into space. If we assume that at the time of the explosion there were equal amount of U-235 and U-238, how long ago did the star(s) explode that released the elements that formed our Earth? The present U-235/U-238 ratio is 0.00700. [The half-lives of U-235 and U-238 are 0.700 × 109 yr and 4.47 × 109 yr.]
Answer:
t = 5.94x10⁹ years.
Explanation:
The time of the explosion can be calculated using the decay equation:
[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]
Where:
N(t): is the quantity of the element at the present time
N(0): is the quantity of the element at the time of explosion
λ: is the decay constant
t: is the time
Knowing that the present U-235/U-238 ratio is 0.00700 and that at the time of the explosion there were equal amount of U-235 and U-238, we have:
[tex]\frac{N_{U-235}}{N_{U-238}} = \frac{N_{U-235_{0}}e^{-\lambda_{U-235} t}}{N_{U-238_{0}}e^{-\lambda_{U-238} t}}[/tex] (1)
The decay constant is equal to:
[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]
For the U-235 we have:
[tex] \lambda_{U-235} = \frac{ln(2)}{0.700 \cdot 10^{9} y} = 9.90 \cdot 10^{-10} y^{-1} [/tex]
For the U-238 we have:
[tex] \lambda_{U-238} = \frac{ln(2)}{4.47 \cdot 10^{9} y} = 1.55 \cdot 10^{-10} y^{-1} [/tex]
By introducing the values of [tex]\lambda_{U-235}[/tex] and [tex]\lambda_{U-238}[/tex] into equation (1) we have:
[tex]0.00700 = \frac{e^{-9.90 \cdot 10^{-10} t}}{e^{-1.55 \cdot 10^{-10} t}}[/tex]
[tex]0.00700 = e^{(-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t}[/tex]
[tex]ln(0.00700) = (-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}) t[/tex]
[tex]t = \frac{ln(0.00700)}{-9.90 \cdot 10^{-10} + 1.55 \cdot 10^{-10}} = 5.94 \cdot 10^{9} y[/tex]
Therefore, the star exploded 5.94x10⁹ years ago.
I hope it helps you!
The explosion of the star that released the uranium that formed Earth happened approximately 6 billion years ago, determined using the half-lives of U-235 and U-238 and the current U-235/U-238 ratio.
The question regarding the age of the elements from a star explosion can be addressed using the concept of radioactive decay and isotope half-lives, specifically of Uranium-235 (U-235) and Uranium-238 (U-238). Using the current U-235/U-238 ratio of 0.00700 and knowing the half-lives of U-235 (0.700 × 109 years) and U-238 (4.47 × 109 years), we can calculate that the star exploded approximately 6 billion years ago based on the change in the U-235/U-238 ratio from an assumed equal initial amount. This estimate is consistent with the age of the solar system and the time it would take for such materials to coalesce into a planetary body like Earth.
A wedge-shaped air film is made by placing a small slip of paper between the edges of two thin plates of glass 12.5 cm long. Light of wavelength 600 nm in air is incident normally on the glass plates. If interference fringes with a spacing of 0.200 mm are observed along the plate, how thick is the paper? This form of interferometry is a very practical way of measuring small thicknesses.
Answer:
The thickness of the paper is [tex]t = 188\mu m[/tex]
Explanation:
From the question we are told that
The length of the wedge-shaped air film [tex]L = 12.5 cm = \frac{12.5}{100} = 0.125m[/tex]
The wavelength of the light is [tex]\lambda = 600nm = 600* 10^{-9}m[/tex]
The spacing of the interference fringe is [tex]D = 0.200mm = \frac{0.200}{1000} = 0.2*10^{-3} m[/tex]
For destructive interference the thickness is mathematically represented as
[tex]t =\frac{\lambda * L}{2 * D }[/tex]
Substituting values
[tex]t = \frac{600 * 10^{-9} * 0.125 }{2 * 0.2 *10^{-3}}[/tex]
[tex]t = 188\mu m[/tex]
A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a tuning fork over the opening. The sound waves that travel down into the tube will reflect from the surface of the water and come back to the tuning fork. This is like a half open pipe with a fixed end condition on the water side and an open end condition on the openThe condition for resonance is if the depth of the air column is an odd integer multiple of lamda/4. The speed of sound is 345 m/s.A) If a tuning fork was used for this experiment, what is the depth of the air column that will satisfy the resonance condition for the fundamental mode?B) What is the depth of L3the air column for the 3rd harmonic resonance?C) Is there a depth that will result in a 2nd harmonic resonance? Explain
Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:
[tex]f_n=\frac{nv_s}{4L}[/tex]
n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:
[tex]f_1=\frac{v_s}{4L}[/tex]
B) For the 3rd harmonic you have:
[tex]f_3=\frac{3v_s}{4L}[/tex]
C) For the 2nd harmonic:
[tex]f_2=\frac{2v_s}{4L}[/tex]
A recent home energy bill indicates that a household used 475475 kWh (kilowatt‑hour) of electrical energy and 135135 therms for gas heating and cooking in a period of 1 month. Given that 1.00 therm is equal to 29.329.3 kWh, how many milligrams of mass would need to be converted directly to energy each month to meet the energy needs for the home?
Answer:
Explanation:
Given that,.
A house hold power consumption is
475 KWh
Gas used is
135 thermal gas for month
Given that, 1 thermal = 29.3 KWh
Then,
135 thermal = 135 × 29.3 = 3955.5 KWh
So, total power used is
P = 475 + 3955.5
P =4430.5 KWh
Since 1 hr = 3600 seconds
So, the energy consumed for 1hr is
1KW = 1000W
P = energy / time
Energy = Power × time
E = 4430.5 KWhr × 1000W / KW × 3600s / hr
E = 1.595 × 10^10 J
So, using Albert Einstein relativity equation
E = mc²
m = E / c²
c is speed of light = 3 × 10^8 m/s
m = 1.595 × 10^10 / (3 × 10^8)²
m = 1.77 × 10^-7 kg
Then,
1 kg = 10^6 mg
m = 1.77 × 10^-7 kg × 10^6 mg / kg
m = 0.177mg
m ≈ 0.18 mg
Total monthly energy consumption of the house in kilowatt-hours (kWh) is calculated. Then, with the help of the equation E = mc², where c is the speed of light, the equivalent mass of energy in kilograms is found by converting energy to Joules and solving for mass.
Explanation:The household uses 475475 kWh of electrical energy and 135135 therms for gas heating and cooking monthly. Considering that 1.00 therm is equal to 29.329.3 kWh, total energy consumption for the month in kWh can be calculated as (475475 + (135135 × 29.3)) kWh.
Einstein’s famous equation E = mc² relating energy (E) and mass (m) allows us to calculate the equivalent mass of this energy.
Here, c is the speed of light. We convert the energy into Joules (1 kWh = 3.6 × 10⁶ J) and then solve for m to find out the mass in kilograms that would need to be converted into energy each month. The question seems to have a typo in the values, you might need to correct them to calculate the accurate amount.
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The light pattern shown is a result from a beam of light being passed through a single slit. The pattern is created from constructive and destructive interference. This phenomenon is due to
A) the wave nature of light.
B) the particle nature of light.
C) the resonance effect of light.
D) the photoelectric effect of light.
Answer:
A
Explanation:
The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of seawater is 1025 kg/m3. (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius = 0.280 m)? (b) For comparison, determine the weight of a jetliner whose mass is 2.44 × 105 kg.
Answer:
A) 27209506.5 N
B) 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
Explanation:
Force du to depth of water is
F = pghA
P = density of salt water = 1025 kg/m3
g = acceleration due to gravity 9.81 m/s2
h = depth of water 11000 m
A = area pressure acts
Area = ¶r^2 = 3.142 x 0.280^2 = 0.246 m^2
Therefore
F = 1025 x 9.81 x 11000 x 0.246
= 27209506.5 N
Weight of a jetliner with mass 2.44 × 10^5 kg is,
2.44×10^5 x 9.81 = 2393640 N
The force on the underwater vehicle is about 11.37 times the weight of the jetliner for comparison.
The force exerted by the water on the sea underwater vehicle's window at the depth of the Mariana Trench is approximately 2.73 x 10⁷ N. For comparison, the weight of a jetliner with a mass of 2.44 x 10⁵ kg is 2.39 x 10⁶ N. Thus, the force exerted on the underwater vehicle at the Mariana's depth is an order of magnitude greater than the weight of the jetliner.
Explanation:To calculate the force that the water exerts on the observation window of an underwater vehicle, you first need to find the pressure at the depth of the Mariana Trench. The pressure is given by the formula P = ρgh, where P is pressure, ρ is density, g is acceleration due to gravity, and h is height (in this case, depth).
Substituting the given values, the pressure at a depth of 11,000 m is about 1.11 x 10⁸ Pa. The force exerted on the window can be found using F = PA, where A is the area of the window. With a radius of 0.280 m, the area of the window (A = πr²) is 0.246 m². Thus, the force applied by the water on the window is F = (1.11 x 10⁸ Pa) (0.246 m²) = 2.73 x 10⁷ N approx.
For the second part of the question, the weight (force) of the jetliner is given by the equation W = m × g. Using the given mass (2.44 x 10⁵ kg) and the acceleration due to gravity (9.8 m/s²), the weight of the jetliner is 2.44 x 10⁵ kg × 9.8 m/s² = 2.39 x 10⁶ N, which is an order of magnitude less than the force exerted on the underwater vehicle at the depth of the Mariana Trench.
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Sandra is riding a bicycle at 10.0 meters/second. She slows down to 2.0 meters/second in 10 seconds.
Answer:
Acceleration= final velocity-initial velocity/time taken
so
a= 10-2/10
a=-8/10
a= -.8 meters per second
Answer: 8.0 = 8 meters/seconds.
Explanation:
Started at= 10.0 meters/seconds.
Slows down= 2.0
10-2=8
Sandra is currently riding her bicycle at 8.0 meters/seconds.
*Mark brainlist* please :))
8. A tuning fork vibrating with a sonometer wire 20cm long produces 5 beats per second. Given
that the beat frequency does not change if the length of the wire is changed to 21cm, calculate
the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm.
Answer:
Explanation:
length of wire is proportional to frquency of sound produced by it.
n₁ /n₂ = l₂ / l₁
= 21 / 20
If n be the frequency of tuning fork
n₁ = n + 5 ; n₂ = n - 5 ( no of beat / s = frequency difference )
n + 5 / n - 5 = 21 / 20
20n + 100 = 21n - 105
n = 205 Hz
frequency of tuning fork = 205
the frequency of the tuning fork and frequency of sonometer wire when changed to 21cm. = n - 5 = 200 Hz .
Given the beat frequency and the inverse relationship between the length of the sonometer wire and frequency, we determine the tuning fork frequency. Beat frequency remains constant despite the change in wire length. Calculation involves solving system of linear equations with fundamentals of wave physics.
To find the frequency of the tuning fork and the frequency of the sonometer wire for both lengths, we utilize the concept of beat frequency. The beat frequency is given by the difference in frequency between the tuning fork and the sonometer wire.
Let’s denote the frequency of the tuning fork as ft and the frequency of the sonometer wire for 20 cm as f1. Given that the beat frequency is 5 Hz, we have:
|ft - f1| = 5 Hz
When the length of the wire is changed to 21 cm, assume the new frequency of the sonometer wire to be f2. Beat frequency remains unchanged:
|ft - f2| = 5 Hz
The frequency of a vibrating string is inversely proportional to its length:
f1 * L1 = f2 * L2
Given L1 = 20 cm and L2 = 21 cm,
f1 * 20 = f2 * 21 which simplifies to:
f2 = (20/21) * f1
Since |ft - f1| = 5 and |ft - f2| = 5, solving these equations requires knowing the possible frequencies of the sonometer.
Determine f1 and f2 using the equations:
Solve both scenarios for ft + 5 = f1 and ft - 5 = f2:
If ft= f1 ± 5 Hz,
Scenario 1: for f1 = (n/20), f2 = (20/21) * (n/20) = n/21, verify both.
Consider three starships that pass by an observer on Earth. Starship A is traveling at speed v=c/3v=c/3 relative to Earth and has a clock placed aboard. Starship B is traveling at speed v=c/3v=c/3 relative to Earth and in the same direction as Starship A. Starship C is traveling at speed v=c/3v=c/3 relative to Earth, but in the opposite direction as Starship A. In which reference frame can a time interval be measured that equals the time interval measured by the clock aboard Starship A?
An observer on Starship B will measure the same time interval as that on Starship A, because they are both travelling at the same speed and in the same direction relative to Earth. Starship C's measurement would differ due to its opposite direction of travel relative to A and B.
Explanation:To determine which reference frame measures a time interval equal to the time interval measured by the clock aboard Starship A, we need to consider the principles of the theory of special relativity. Since Starship B is traveling at the same speed and in the same direction as Starship A relative to Earth, the time dilation effect will be the same for both starships. Therefore, an observer on Starship B will measure the same time interval on their clock as that measured on Starship A's clock.
In contrast, Starship C is moving at the same speed but in the opposite direction relative to Earth, and its relative velocity to Starships A and B is not zero. As a result, the time interval measured on Starship C will not match the time interval measured on Starship A's clock.
So, the time interval that equals the time interval measured by the clock aboard Starship A can be measured in the reference frame of Starship B, which is moving at the same speed and in the same direction as Starship A relative to Earth.
A parallel-plate capacitor of capacitance 20 µF is fully charged by a battery of 12 V. The battery is then disconnected. A dielectric slab of K = 4 is slipped between the two plates of the capacitor:
(a) Find the change in potential energy of the capacitor.
(b) Does the potential energy increase or decrease? Explain
Answer:
Explanation:
capacitance = 20 x 10⁻⁶ F .
potential V = 12 V
charge = CV
= 20 x 10⁻⁶ x 12
Q = 240 x 10⁻⁶ C
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 20 x 10⁻⁶
= 1440 x 10⁻⁶ J
b )
In this case charge will remain the same but capacity will be increased 4 times
new capacity C = 4 x 20 x 10⁻⁶
= 80 x 10⁻⁶
energy = Q² / 2C
= (240 x 10⁻⁶ )² / 2 x 4 x 20 x 10⁻⁶
= 360 x 10⁻⁶ J .
potential energy will decrease from 1440 x 10⁻⁶ J to 360 x 10⁻⁶ J
[03.02]
If you were going to describe the relationship between current, voltage, and power, you could say: (1 point)
Group of answer choices
"If there is a decrease in power, there must have been an increase in either current or voltage."
"If you decrease your voltage, and all other factors remain the same, you will have an increase in power."
"In order to gain more power you would need to increase either current or voltage."
"The only way to increase your power is to change your voltage and keep your current the same."
Answer:
in other to gain more power you would need to increase either current or voltage
Explanation:
In other to gain more power, you would need to increase either current or voltage,because power is directly proportional to voltage when current is kept constant.also power is directly proportional to current when voltage is kept constant
Which pair shows the law of reflection?
Answer:
The answer is A and C .
Explanation:
Reflection of object is reflected through the Normal (mirror) .
*incident angle = refracted angle
A solid copper sphere hangs at the bottom of a steel wire of negligible mass. The top end of the wire is fixed. When the wire is struck, it vibrates with a fundamental frequency of 300 Hz. The copper object is then submerged in water so that half its volume is below the water line. Determine the new fundamental frequency. The density of water, copper, and steel is 1000kg/m³, 8960kg/m³, and 8050kg/m³ respectively
Answer:
291.509 Hz
Explanation:
Fundamental frequency, often referred to simply as the fundamental, is defined as the lowest frequency of a periodic waveform.
It is a vital concept in musical instruments and many aspects .
See the attached file for the solution to the given problem.
Buffy is rolling along in her 10.4 kg wagon at 3.4 m/s (in the positive direction) when she jumps off the back. She continues to move forward at 1.7 m/s relative to the ground. This causes her wagon to go speeding forward at 8.04 m/s relative to the ground. How much does Buffy weigh
Answer:
Buffy weigh's 28.39 kg
Explanation:
Given;
mass of Buffy and wagon, M = 10.4 kg
final velocity of Buffy - wagon system, v = 3.4 m/s
Buffy's velocity relative to the ground, u₁ = 1.7 m/s
Wagon's velocity relative to the ground, u₂ = 8.04 m/s
Buffy's velocity = 1.7 - 3.4 = - 1.7 m/s
Wagon's velocity = 8.04 - 3.4 = 4.64 m/s
Apply the principle of conservation of linear momentum;
1.7 x m = 4.64 x 10.4
where;
m is mass of wagon
1.7 m = 48.256
m = 48.256 / 1.7
m = 28.39 kg
Therefore, Buffy weigh's 28.39 kg
If a light bulb is missing or broken in a parallel circuit, will the other light bulb right ?
Outside my window a squirrel is scurrying up and down a tree. Its position function is given by s(t) = t 3 − 12t 2 + 36t for the seven seconds that I’m watching it (so from t = 0 to t = 7). (a) What is the velocity function, v(t), for the motion of the squirrel? (b) What is the acceleration function, a(t), for the motion of the squirrel? (c) At the four second mark, is the squirrel moving up the tree or down the tree? Justify your answer.
Answer:
Explanation:
Given
Position of squirrel is given by
[tex]s(t)=t^3-12t^2+36t[/tex]
Velocity is given by
[tex]v(t)=\frac{ds(t)}{dt}=\frac{d(t^3-12t^2+36t)}{dt}[/tex]
[tex]v(t)=3t^2-12\times 2t+36[/tex]
[tex]v(t)=3t^2-24t+36[/tex]
(b) acceleration is given by
[tex]a(t)=\frac{da(t)}{dt}=\frac{d(3t^2-24t+36)}{dt}[/tex]
[tex]a(t)=6t-24[/tex]
(c)at [tex]s(3)=3^3-12(3)^2+36(3)[/tex]
[tex]s(3)=27\ m[/tex]
at [tex]s(4)=4^3-12(4)^2+36(4)[/tex]
[tex]s(4)=16\ m[/tex]
at [tex]t=3\ s[/tex] Position is [tex]27\ m[/tex] and at [tex]t=4\ s[/tex] position is [tex]16\ m[/tex]
therefore squirrel is moving down
How much heat is required, in joules, to complete the of 15.0g of water’s phase change from liquid to steam?
A. 150.7 J
B. 33,900 J
C. 62.7 J
D. 5,010 J
Answer: 62.7
Explanation:
The specific heat capacity of water is 4.186 J you multiply 4.186 and 15.0 and you will get 62.7
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in series. Initially, the switch is open and has been open for a very long time. At time t = 0 s, the switch is suddenly closed. How long after closing the switch will the potential difference across the inductor be 24 V?
Answer:
1.6 s
Explanation:
To find the time in which the potential difference of the inductor reaches 24V you use the following formula:
[tex]V_L=V_oe^{-\frac{Rt}{L}}[/tex]
V_o: initial voltage = 60V
R: resistance = 24-Ω
L: inductance = 42H
V_L: final voltage = 24 V
You first use properties of the logarithms to get time t, next, replace the values of the parameter:
[tex]\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s[/tex]
hence, after 1.6s the inductor will have a potential difference of 24V
To find the time when the potential difference across the inductor will be 24 V in an LR circuit, we need to use the formula for the time-dependent current in an RL circuit, and through mathematical manipulations, substitute into the equation for the induced emf in the inductor and solve for time.
Explanation:The time for the potential difference across the inductor to be 24V in an LR circuit can be calculated using the formula for the time-dependent current in an RL circuit, which accounts for how current evolves over time. According to Faraday's law, a changing current in an inductor generates an opposition voltage, its magnitude is determined by L (inductance of the inductor) times dI/dt (the rate of change of current).
In our case we have the equation for the current in an RL circuit when turned on: I(t) = V/R*(1 - e^(-R*t/L)) and the induced emf in the inductor V(L) = L*dI/dt. Our goal is to find the time when V(L) equals 24V. This requires substituting I(t) into the equation for V(L), setting V(L) equal to 24V, and solving for t.
This process involves the principles of RL circuits, the characteristic time constant, and Faraday's law.
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A 100-g toy car moves along a curved frictionless track. At first, the car runs along a flat horizontal segment with an initial velocity of 3.33 m/s. The car then runs up the frictionless slope, gaining 0.108 m in altitude before leveling out to another horizontal segment at the higher level. What is the final velocity of the car if we neglect air resistance
Answer:
2.994 m/s
Explanation:
m = Mass of the car = 100 g
[tex]v_1[/tex] = Initial velocity = 3.33 m/s
h = Height = 0.108 m
g = Acceleration due to gravity = 9.81 m/s²
We know that energy in the system is conserved so
[tex]\dfrac{1}{2}mv_1^2=mgh+\dfrac{1}{2}mv_f^2\\\Rightarrow v_f=\sqrt{2\left({\dfrac{1}{2}v_1^2-gh}\right)}\\\Rightarrow v_f=\sqrt{2\left(\dfrac{1}{2}3.33^2-9.81\times 0.108\right)}\\\Rightarrow v_f=2.994\ m/s[/tex]
The final velocity of the car is 2.994 m/s
The final velocity of the car is approximately 6.05 m/s.
Explanation:To find the final velocity of the car, we can use the principles of conservation of energy. In this case, the initial kinetic energy of the car is equal to its final potential energy. The initial kinetic energy can be calculated using the formula K.E. = 1/2 mv^2, where m is the mass of the car (100 g = 0.1 kg) and v is the initial velocity (3.33 m/s).
The final potential energy can be calculated using the formula P.E. = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the change in altitude (0.108 m).
Setting the initial kinetic energy equal to the final potential energy, we can solve for the final velocity of the car:
1/2 mv^2 = mgh
Substituting the given values, we get:
1/2 (0.1 kg)(v^2) = (0.1 kg)(9.8 m/s^2)(0.108 m)
Simplifying and solving for v, we find that the final velocity of the car is approximately 6.05 m/s.
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A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of 0.9930c. An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?
Answer:
Her current biological age = 24.02 years
Explanation:
From time dilation equation, we know that;
t = t_o * [√(1-(v²/c²))]•L_o
Where;
t = dilated time
t_o = stationary time
v = the speed of the moving object
c = the speed of light in a vacuum
First, let's convert the rest time (t_o) from light years to years.
Thus;
t_o = [c/0.993c] * [42.2]
c will cancel out and we now have;
t_o = 42.5 years
Since t = t_o * [√(1-(v²/c²))]
Thus; t = 42.5 * [√(1-(0.993²c²/c²))]
t = 42.5/[√(1 - (0.993²))]
t = 42.5 * 0.1181
t = 5.02 years
Since the astronaut was 19 years old when the probe left the earth, thus;
Her current biological age now the probe has reached Capella, will be;
19 + 5.02 = 24.02 years
The number of protons in the nucleus of an atom determines the species of the atom, i.e., the element to which the atom belongs. An atom has the same number of protons and neutrons. But the electron number cannot be used instead because
A. electrons are not within the nucleus
B. electrons are negatively charged
C. electrons can be removed from or added to an atom
D.electrons are lighter than protons
The number of protons in the nucleus of an atom determines the element that the atom belongs to.
Explanation:The subject of this question is Chemistry. The number of protons in the nucleus of an atom determines the species or element to which the atom belongs. This is because each element has a unique number of protons, known as the atomic number. The other options are not correct because:
Electrons are not within the nucleus: Electrons are found in electron shells outside the nucleus of an atom.Electrons are negatively charged: While electrons are negatively charged, this does not affect their use in determining the species of an atom.Electrons can be removed from or added to an atom: While electrons can be added or removed from an atom, their number does not define the species of the atom.Electrons are lighter than protons: Although electrons are lighter than protons, this is not the reason why their number cannot be used to determine the species of an atom.Learn more about protons in the nucleus here:https://brainly.com/question/30654172
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Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large. Write a differential equation that expresses Newton's Law of Cooling for this particular situation. (Use t as the independent variable, y as the dependent variable, R as the room temperature, and k as a proportionality constant.)
Answer:
[tex]\frac{dy}{dt}=k(y_t-R)[/tex]
Explanation:
According to Newton’s law of cooling, the rate of loss of heat from a body and the difference in the temperature of the body and its surroundings are proportional to each other.
[tex]\frac{dy}{dt}=k(y_t-R)[/tex]
Here, [tex]y_t[/tex] represents temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.
The equation that represents Newton's Law of Cooling for this particular situation is dy/dt = k(yt-R)
What is Newton’s law?
According to Newton’s law of cooling refer, the rate of loss of heat from a body and also the difference in the temperature of the body and also its surroundings are proportional to each other.
dy/dt is = k(yt-R)
Therefore, yt conveys temperature at time t, R as the room temperature, t as the independent variable, y as the dependent variable.
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Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+β .
To evaluate the final kinetic energy of a supply spacecraft under the given tractor beam force, you would need to integrate the force over the displacement. Direct calculation requires specific values for α, β, and displacement x, which are not provided.
Explanation:The question involves calculating the final kinetic energy of a supply spacecraft under a specific tractor beam force, F(x)=αx3+β. This calculation would typically require the integration of the force over the displacement, as kinetic energy can be evaluated through the work-energy principle where work done by a force in moving an object is equal to the change in kinetic energy of the object. Here, without specific values or further context provided for α, β, or the displacement, x, a direct calculation cannot be made. However, in physics, especially in the study of mechanics and spacecraft dynamics, understanding how forces affect motion and energy forms the basis for analyzing and optimizing space missions. Evaluating the final kinetic energy would typically involve integrating the force function over the spacecraft's path, considering initial conditions, and any external forces or resistances.
An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick. What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, m proton = 1.67 × 10-27 kg, = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s) a. 7.7 × 10-10 % b. 0.11% c. 1.1% d. 0.011% e. 1.1 × 10-4 %
Answer:
d) 0.011%
Explanation:
The probability for tunneling the barrier is given by the following formula:
[tex]P=exp(-2d\sqrt{\frac{2m_e(U_o-E)}{\hbar ^2} }\ )[/tex] ( 1 )
me: mass of the electron
Uo: energy of the barrier
E: energy of the electron
d: thickness of the barrier
By replacing the values of the parameters in (1), you obtain:
[tex]P=exp(-2(0.20*10^{-9}m)\sqrt{\frac{2(9.11*10^{-31}kg)(100eV-80eV)(1.60*10^{-19}J)}{(1.055*10^{-34}Js)^2}})\\\\P=e^{-9.15}=1.08*10^{-4}\approx0.011\%[/tex]
hence, the probability is 0.011% (answer d)
A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed
Answer:
Explanation:
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
The wok done by force will be converted into rotational kinetic energy
F x d = 1/2 I ω²
F is force applied , d is displacement , I is moment of inertia of disc and ω
is angular velocity of disc
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
The angular speed should be 17.18 rad / s
Calculation of the angular speed:Since
moment of inertia of the disc I = 1/2 m r²
= .5 x 8 x .25²
= .25 kg m²
Now the work done by force should be converted into the rotational kinetic energy
F x d = 1/2 I ω²
here,
F is the force applied,
d is displacement,
I is moment of inertia of disc
and ω is angular velocity of disc
So,
41 x .9 = 1/2 x .25 ω²
ω² = .25
ω = 17.18 rad / s
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One particle has a mass of 3.12 x 10-3 kg and a charge of +8.8 C. A second particle has a mass of 7.1 x 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.15 m, the speed of the 3.12 x 10-3 kg particle is 131 m/s. Find the initial separation between the particles.
Answer:[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Explanation:
Given
mass of first particle is [tex]m_1=3.12\times 10^{-3}\ kg[/tex]
mass of second particle is [tex]m_2=7.1\times 10^{-3}\ kg[/tex]
Charge on both the particle [tex]q=8.8\times 10^{-6}\ C[/tex]
Now final speed of first particle is [tex]v_1=131\ m/s[/tex]
Final separation between particles is [tex]r=0.15\ m[/tex]
As there is no external force therefore linear momentum is conserved
[tex]0+0=m_1v_1+m_2v_2[/tex]
[tex]0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2[/tex]
[tex]v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131[/tex]
[tex]v_2=-57.56\ m/s[/tex]
Conserving total energy
Initial Kinetic energy +Initial Potential energy=Final Kinetic energy +Final Potential energy
[tex]\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}[/tex]
[tex]\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646[/tex]
[tex]\Rightarrow \frac{0.696}{r_i}=43.177[/tex]
[tex]r_i=0.016119\ m\approx 16.119\ mm[/tex]
Using the law of conservation of linear momentum and conservation of total energy, the obtained initial separation between the particles is 0.01612 m.
Conservation of Linear MomentumGiven that the masses of the two particles are;
[tex]m_1 = 3.12\times 10^{-3}\,kg\\m_2 = 7.1\times 10^{-3}\,kg[/tex]
Also, the charges of both the particles are equal.
[tex]q_1 = q_2 = +8.8\times 10^{-6} \,C[/tex]
The final separation between the particles is;
[tex]r_f = 0.15\,m[/tex]
Also, the final speed of the first particle is;
[tex](v_1)_{final} = 131\,m/s[/tex]
There is no external force applied here; so by the law of conservation of linear momentum, we have;
[tex](m_1v_1)_{initial} +(m_2v_2)_{initial}=(m_1v_1)_{final} +(m_2v_2)_{final}[/tex]
But initially, the particles are at rest, so the initial velocities are zero.
[tex]0+0=(3.12\times 10^{-3}\,kg \, \times 131\,m/s ) +(7.1\times 10^{-3}\,kg\,)\,(v_2)_{final}\\[/tex]
[tex]\implies (v_2)_{final}=-\frac{3.12\times 10^{-3}\,kg \, \times 131\,m/s}{7.1\times 10^{-3}\,kg} =-57.57\,m/s[/tex]
Conservation of EnergyNow, applying the law of conservation of total energy, we get;
[tex](KE)_{\,initial}\, + \,(PE)_{\,initial} = (KE)_{\,final}\, + \,(PE)_{\,final}[/tex]
But initially, the particles are at rest; so they have no initial kinetic energy.
They have electrostatic potential alone initially.
[tex]0+k\frac{q^2}{r_i} = \frac{1}{2}(m_1 v_1)_{initial} + \frac{1}{2}(m_2 v_2)_{final} + k\frac{q^2}{r_f}[/tex]
Substituting the known values, we get;
[tex](9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_i} =[ \frac{1}{2}(3.12\times 10^{-3}\,kg )\times (131\,m/s)^2] + [\frac{1}{2}(7.1\times 10^{-3}\,kg) \times (-57.57\,m/s)^2 + (9\times 10^9 \, Nm^2/C^2)\frac{(8.8\times 10^{-6} \,C)^2}{r_f}[/tex]
[tex]\implies\frac{0.696\,Nm^2}{r_i} =26.77\,J+11.76\,J+ 4.64\,J[/tex]
[tex]\implies r_i =\frac{0.696\,Nm^2}{43.17\,J}=0.01612\,m[/tex]
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You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion. What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path? If the string's maximum tension is 45 N, what is the maximum speed the rock can have so that the string does not break? At what point in the vertical circle does this maximum value occur?
Answer:
v (minimum speed) = 2.90 m/sec.
[tex]\\ \\ maximum speed (v)= 6.57 m/sec.\\[/tex]
Maximum value of speed will occur at lowest point of vertical circle.
Explanation:
a) What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?
Using the force balance expression at the top of the circle,
Gravitational Force + Tension force = Centrifugal force
[tex]m*g + T = m*v^2/R[/tex]
Given that : T = 0
R = length of string = 0.86 m
mass of the spinning rock = 0.75 kg
[tex]v = \sqrt{g*R}[/tex]
[tex]v = \sqrt{9.81*0.86}[/tex]
v (minimum speed) = 2.90 m/sec.
b) what is the maximum speed the rock can have so that the string does not break?
Here the force balance at bottom of circle is represented by the illustration:
[tex]T = m*g + m*v^2/R[/tex]
Given that:
maximum tension T = 45 N
maximum speed v = ??
mass m = 0.75 kg
∴
[tex]45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\[/tex]
c)
At what point in the vertical circle does this maximum value occur?
Maximum value of speed will occur at lowest point of vertical circle.
This is so because at the lowest point; the tension in string will be maximum.
When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or P, wave has a speed of about 8.0 km/s and the secondary, or S, wave has a speed of about 4.5 km/s. A seismograph, located some distance away, records the arrival of the P wave and then, 77.2 s later, records the arrival of the S wave. Assuming that the waves travel in a straight line, how far (in terms of m) is the seismograph from the earthquake?
Answer:[tex]d=7.94\times 10^5\ m[/tex]
Explanation:
Given
Speed of Primary wave [tex]v_1=8\ km/s[/tex]
Speed of secondary wave [tex]v_2=4.5\ km/s[/tex]
difference in timing of two waves are [tex]77.2\ s[/tex]
Suppose both travel a distance of d km then
[tex]t_1=\frac{d}{8}\quad \ldots (i)[/tex]
[tex]t_2=\frac{d}{4.5}\quad \ldots (ii)[/tex]
Subtract (ii) from (i)
[tex]\frac{d}{4.5}-\frac{d}{8}=77.2[/tex]
[tex]d[\frac{1}{4.5}-\frac{1}{8}]=77.2[/tex]
[tex]d[0.0972]=77.2[/tex]
[tex]d=794.23\ km[/tex]
[tex]d=7.94\times 10^5\ m[/tex]
Final answer:
To calculate the distance to the earthquake's epicenter, we use the time difference between the arrival of P-waves and S-waves and their speeds. By setting up an equation and solving for the distance, the seismograph is found to be approximately 618,940 meters from the epicenter.
Explanation:
When an earthquake occurs, two types of waves are generated: P-waves (primary waves) and S-waves (secondary waves), each with distinctive speeds. To determine the distance to the epicenter of the earthquake, we use the formula d = v × t, where d is distance, v is velocity, and t is time. Given that the P-wave has a speed of 8.0 km/s and the S-wave has a speed of 4.5 km/s, and the time difference of arrival between the two waves is 77.2 seconds, we can calculate the distance from the seismograph to the earthquake's epicenter.
Let the distance be d, then:
Time for P-wave to travel d: d / 8.0 km/sTime for S-wave to travel d: d / 4.5 km/sThe difference in travel time is 77.2 s, so: d / 4.5 km/s - d / 8.0 km/s = 77.2 sTo find the distance d, we solve the equation:
8.0×d - 4.5×d = 77.2 s × (8.0 km/s × 4.5 km/s)3.5×d = 77.2 s × 36 km²/s²d ≈ 618.94 kmNow, to convert kilometers to meters:
d ≈ 618.94 km × 1,000 m/kmd ≈ 618,940 mTherefore, the seismograph is approximately 618,940 meters from the earthquake's epicenter.
In a double‑slit interference experiment, the wavelength is λ = 452 nm λ=452 nm , the slit separation is d = 0.190 mm d=0.190 mm , and the screen is D = 49.0 cm D=49.0 cm away from the slits. What is the linear distance Δ x Δx between the eighth order maximum and the third order maximum on the screen?
Answer:
Δx = 5.82mm
Explanation:
To find the distance between the eight maximum and the third one you use the following formula:
[tex]x_m=\frac{m\lambda D}{d}[/tex] (1)
λ: wavelength = 452*10^-9 m
m: order of the fringes
D: distance to the scree = 0.49m
d: distance between slits = 0.190*10^-3 m
you use for m=8 and m=3, then you calculate x8 - x3:
[tex]x_8=\frac{8(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=9.32*10^{-3}m\\\\x_3=\frac{3(452*10^{-9}m)(0.49m)}{0.190*10^{-3}m}=3.49*10^{-3}m\\\\\Delta x_{8,3}=5.82*10^{-3}m=5.82mm[/tex]
hence, the distance between these fringes is 5.82mm
The density of mercury is 13.546 g/cm3 . Calculate the pressure exerted by a column of mercury 76 cm high. Give your answer in Pascals and lbf/in2 . 2. The density of water is 62.43 lbm/ft3 . Calculate the pressure exerted by a column of water 25 ft high. Give your answer in Pascals and lbf/in2 3. What is the power required to pump 10 kg/s water from a height of 5 meters to a height of 30 meters? Report the power in Watts and hp. 4. Water is being pumped by the application of pressure at point 1, see below, up to a height of 50 m at a rate of 1 kg/s. At the top, point 2 the pressure is nearly one atmosphere (105 N/m2 )
Answer:
The required solution is 100890 Pa and 14.3lb/in²
Explanation:
See attached image
Answer:
1. p = 14.63 lb/in² or 100890.608 Pa
2. p = 74676 Pa or 10.83 lb/in²
3. P = 2450 W or 3.28 hp
4. [tex]p_{1}[/tex] = 490105 N/m²
Explanation:
1. Let's begin by listing out the given parameters:
density of mercury = 13.546 g/cm³ = 13546 kg/m³,
height of column = 76 cm = 0.76 m, acceleration due to gravity = 9.8m/s²
Using Pressure = density * acceleration due to gravity * height of column
p = ρ g h = 13546 * 9.8 * 0.76
p = 100890.608 Pa
To get the answer in lb/in², divide by 6895
p = 100890.608 ÷ 6895 = 14.632
p = 14.63 lb/in²
2. Let's list out the parameters given:
density of water = 62.43 lbm/ft³ = 62.43 * 16.018 = 1000kg/m³,
height of column = 25 ft = 25 ÷ 3.281 = 7.62 m,
acceleration due to gravity = 9.8m/s²
Using Pressure = density * acceleration due to gravity * height of column
p = ρ g h = 1000 * 9.8 * 7.62
p = 74676 Pa
To convert from Pa to lb/in², divide by 6895
p = 74676 ÷ 6895
p = 10.83 lb/in²
3. Let's list out the parameters given:
mass flow rate (ṁ) = 10 kg/s, [tex]h_{1}[/tex] = 5 m, [tex]h_{2}[/tex] = 30 m, Δh = 30 - 5 = 25 m, g = 9.8 m/s²
Using Power = Energy (Potential Energy) ÷ Time
Energy (Potential Energy) = m g h
Power = mgΔh ÷ t; m÷ t = ṁ
Substitute ṁ into the equation
P = ṁ g h = 10 * 9.8 * 25
P = 2450 W
To convert from W to hp, divide by 746
P = 2450 ÷ 746 = 3.284
P = 3.28 hp
4. Let's list out the parameters given:
height (Δh) = 50 m, ṁ = 1 kg/s, g = 9.8 m/s²,
p2 = 105N/m², ρ = 1000 kg/m³
Using Bernoulli's Equation,
p1 + ½ρ([tex]v_{1}[/tex])² + ρgh1 = p2 + ½ρ([tex]v_{2}[/tex])² + ρgh2
Assuming steady state flow; [tex]v_{2}[/tex] = [tex]v_{1}[/tex] ⇒ [tex]v_{2}[/tex] - [tex]v_{1}[/tex] = 0
[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ½ρ([tex]v_{2}[/tex] - [tex]v_{1}[/tex])² + ρg([tex]h_{2}[/tex] - [tex]h_{1}[/tex])
[tex]p_{1}[/tex] - [tex]p_{2}[/tex] = ρgΔh
[tex]p_{1}[/tex] - 105 = 1000 * 9.8 * 50
[tex]p_{1}[/tex] = 490000 + 105 = 490105
[tex]p_{1}[/tex] = 490105 N/m²