Answer:
Explanation:
The problem relates to Compton Effect in which electrons are scattered due to external radiation . The electron is scattered out and photons relating to radiation also undergo scattering at angle θ .
The formula relating to Compton Effect is as follows
[tex]\lambda_f-\lambda_i=\frac{h}{m_0c} (1-cos\theta)[/tex]
Here [tex]\lambda_i[/tex] = 3 0 x 10⁻¹¹
For longest [tex]\lambda_f[/tex] θ =180°
[tex]\lambda_f[/tex] = [tex]\lambda_i + \frac{2\times h}{m_0c}[/tex]
= .3 x 10⁻⁹ + [tex]\frac{2\times6.6\times 11^{-34}}{9\times10^{-31}\times3\times10^8}[/tex]
= .348 nm
For shortest wavelength θ = 0
Putting this value in the given formula
[tex]\lambda_f=\lambda_i[/tex]
[tex]\lambda_f[/tex] = .3 nm
A solid cylinder and a thin-walled hollow cylinder have the same radii. They are rolling horizontally toward the bottom of an incline. Assuming there is no friction, what is the maximum height each one reaches (in terms of the variables involved)?
Answer:
height solid / height hollow = 3/4
Explanation:
the height reached is directly proportional to the original KE of the object.
Total KE solid
=[tex]1/2mv^2+1/2\times1/2m r^2(v^2 / r^2 )=3/4mv^2[/tex]
Total KE hollow = [tex]1/2mv^2+1/2mr^2(v^2 / r^2 )=mv^2[/tex]
height solid / height hollow = 3/4
A spaceship of mass 8600 kg is returning to Earth with its engine turned off. Consider only the gravitational field of Earth. Let R be the distance from the center of Earth. In moving from position R1 = 7300 km to position R1 = 6700 km the kinetic energy of the spaceship increases by _____.
The kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.
Explanation:To find the change in kinetic energy of the spaceship as it moves from position R1 = 7300 km to position R1 = 6700 km in the gravitational field of Earth, we can use the gravitational potential energy and the fact that the mechanical energy is conserved. The change in kinetic energy is equal to the negative of the change in potential energy.
The gravitational potential energy (U) is given by the formula:
U = -G * (m1 * m2) / R
Where:
- G is the universal gravitational constant (approximately 6.674 × 10^(-11) N·(m/kg)^2).
- m1 is the mass of the Earth (approximately 5.972 × 10^24 kg).
- m2 is the mass of the spaceship (8600 kg in this case).
- R is the distance from the centre of the Earth.
Now, let's calculate the potential energy at R1 = 7300 km and R2 = 6700 km:
For R1 = 7300 km:
R1 = 7300 km = 7,300,000 meters
U1 = -G * (m1 * m2) / R1
U1 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 7,300,000 m
U1 ≈ -5.125 × 10^9 J
For R2 = 6700 km:
R2 = 6700 km = 6,700,000 meters
U2 = -G * (m1 * m2) / R2
U2 = -6.674 × 10^(-11) N·(m/kg)^2 * (5.972 × 10^24 kg * 8600 kg) / 6,700,000 m
U2 ≈ -5.504 × 10^9 J
Now, let's find the change in potential energy (ΔU) as the spaceship moves from R1 to R2:
ΔU = U2 - U1
ΔU ≈ (-5.504 × 10^9 J) - (-5.125 × 10^9 J)
ΔU ≈ -0.379 × 10^9 J
So, the change in potential energy is approximately -0.379 × 10^9 Joules.
The change in kinetic energy is equal to the negative of the change in potential energy, so:
Change in kinetic energy = -ΔU
Change in kinetic energy ≈ 0.379 × 10^9 J
Therefore, the kinetic energy of the spaceship increases by approximately 0.379 × 10^9 Joules as it moves from R1 = 7300 km to R2 = 6700 km in the gravitational field of Earth.
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In simple harmonic motion, the speed is greatest at that point in the cycle whenA) the magnitude of the acceleration is a maximum.B) the displacement is a maximum.C) the magnitude of the acceleration is a minimum.D) the potential energy is a maximum.E) the kinetic energy is a minimum.
Answer:
C) the magnitude of the acceleration is a minimum.
Explanation:
As we know that ,the general equation of the simple harmonic motion given as
The displacement x given as
x=X sinω t
Then the velocity v will become
v= X ω cosωt
The acceleration a
a= - X ω² sinω t
The speed of the particle will be maximum when cosωt will become 1 unit.
It means that sinωt will become zero.So acceleration and displacement will be minimum.
Therefore when speed is maximum then acceleration will be minimum.
At the mean position the speed of the particle is maximum that is why kinetic energy also will be maximum and the potential energy will be minimum.
Therefore option C is correct.
In simple harmonic motion, the speed is greatest when the displacement is a maximum.
Explanation:In simple harmonic motion, the speed is greatest at the point in the cycle when the displacement is a maximum. When an object is at its maximum displacement, it has the maximum potential energy, and as it moves towards the equilibrium position, the potential energy is converted into kinetic energy, leading to an increase in speed.
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In an oscillating L C circuit, the maximum charge on the capacitor is 1.5 × 10 − 6 C and the maximum current through the inductor is 5.5 mA.
(a) What is the period of the oscillations?
(b) How much time elapses between an instant when the capacitor is uncharged and the next instant when it is fully charged?
Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Explanation:
First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C
a) I(t) = -ωqosin(ωt+φ)
⇒Io= ωqo
⇒ω= Io/qo
also we know that T= 2π/ω
⇒T= [tex]2\pi\frac{q_0}{I_0}[/tex]
now putting the values we get
= [tex]2\pi\frac{1.5\times10^{-6}}{5.5\times10^{-3}}[/tex]
= 1.71×10^{-3} sec
b) note that the time [tex]t_{fc}[/tex] it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is
[tex]t_{fc}= \frac{T}{4}[/tex]
[tex]t_{fc}= \frac{ 1.71\times10^{-3} }{4}[/tex]
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
The most distant galaxies observed to date have a redshift of approximately z =10. Part A How does the wavelength of light we observe from those galaxies compare with its original wavelength when it was emitted? Express your answer as an integer.
To solve the problem it is necessary to apply the concepts related to wavelength and error calculation.
The calculation of the error is usually defined by
[tex]z = \frac{\lambda_o-\lambda_r}{\lambda_r}[/tex]
Where
[tex]\lambda_o =[/tex] Wavelength from the observer
[tex]\lambda_r =[/tex] Wavelength from the rest
The previous formula is exactly equal to,
[tex]z= \frac{\lambda_o}{\lambda_r}-1[/tex]
[tex](z+1) = \frac{\lambda_o}{\lambda_r}[/tex]
[tex]\frac{\lambda_o}{\lambda_r}=(10+1)[/tex]
[tex]\lambda_o=11 \lambda_r[/tex]
Therefore the observed wavelength will be 11 times longer that the emitted wavelength.
If we make the comparison for the ultraviolet region (from 10nm to 400nm) we get that,
[tex]\lambda_o=11 \lambda_r[/tex]
[tex]\lambda_o=11 (\frac{10nm*400nm}}{2})[/tex]
[tex]\lambda_o = 2255nm[/tex]
That is the infrared region of electromagnetic spectra.
The wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.
The redshiftThe redshift (z) corresponds to a change in the way in which the frequency of light photon waves is observed in the spectroscope as a function of the relative velocity between the emitting source and the observing receiver (a measure of distance, of according to cosmological expansion, based on recessional velocity).
A decrease in frequency occurs as the distance from the source galaxy increases, causing the light ray to be captured as a standard color. The deviation or distance measurement is represented by the letter z.
With this information, we can conclude that the wavelength of light we observe from these galaxies compares to its original wavelength when it was emitted, a function of the relative velocity between the emitting source and the observing receiver.
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A college friend of yours who has been postponing taking any science courses hears you talking about the generation of nuclear energy in the Sun and makes the following observation: "The whole idea of the atomic nucleus is pretty ridiculous. If an oxygen nucleus consists of eight protons and eight neutrons, the charge on that nucleus is positive. Since even I learned in high school that like charges repel, such a nucleus would find all its positive protons repelling and quickly fall apart." How would you answer his argument?
Answer:
It's held together by the nuclear force.
Explanation:
There are more elemental forces than just the electromagnetic one. In this case, it is the nuclear force (called also strong force) the one that holds the nucleus together because it is stronger than the electromagnetic force over such short distances as the one inside the atomic nucleus.
Answer:
The strong nuclear force, which is attractive over short distances like the nucleus, and stronger than electricity, holds the nucleus together.
Explanation:
When light with a frequency f1 = 547.5 THz illuminates a metal surface, the most energetic photoelectrons have 1.260 x 10^-19 J of kinetic energy. When light with a frequency f2 = 738.8 THz is used instead, the most energetic photo-electrons have 2.480 x 10^-19 J of kinetic energy
Using these experimental results, determine the approximate value of Planck's constant.
Express your answer using four significant figures.
Answer:
[tex]h=6.377\times10^{-34}kgm^2/s[/tex]
Explanation:
The maximum kinetic energy of the photoelectrons is given by the formula [tex]K_M=hf-\phi[/tex].
We have two situations where for [tex]f_1=547.5\times10^{12}Hz[/tex] we get [tex]K_{M1}=1.26\times10^{-19}J[/tex] and for [tex]f_2=738.8\times10^{12}Hz[/tex] we get [tex]K_{M2}=2.48\times10^{-19}J[/tex], so we have:
[tex]K_{M1}=hf_1-\phi[/tex]
[tex]K_{M2}=hf_2-\phi[/tex]
We can eliminate [tex]\phi[/tex] by substracting the first equation to the second:
[tex]K_{M2}-K_{M1}=hf_2-\phi-(hf_1-\phi)=h(f_2-f_1)[/tex]
Which means:
[tex]h=\frac{K_{M2}-K_{M1}}{f_2-f_1}=\frac{2.48\times10^{-19}J-1.26\times10^{-19}J}{738.8\times10^{12}Hz-547.5\times10^{12}Hz}=6.377\times10^{-34}kgm^2/s[/tex]
On Earth, froghoppers can jump upward with a takeoff speed of 2.8 m/s. Suppose you took some of the insects to an asteroid. If it is small enough they can jump free of it and escape into space. (a) What is the diameter (in km) of the largest spherical asteroid from which they could jump free? Assume typical asteroid denisty of 2.0 g/cm^3. (b) Suppose that one of the froghoppers jumped horizontally from a small hill on an asteroid. What would the diameter (in km) of the ateroid need to be so that the insect could go into a circular orbit just above the surace?
To develop this problem it is necessary to apply the concepts related to kinetic energy and gravitational potential energy.
By conserving energy we know that
[tex]PE = KE[/tex]
[tex]\frac{1}{2}mv^2=\frac{GMm}{r}[/tex]
Where,
m = mass
v = Velocity
G = Gravitational universal constant
M = Mass of Spherical asteroid
m = mass of object
r = Radius
Mass of a Sphere can be expressed as,
[tex]M= \rho* (\frac{4}{3}\pi r^3 )[/tex]
Replacing we have that,
[tex]\frac{1}{2}*2.8^2 - 6.67*10^{-11}*\frac{4\pi*2000*r^3}{3r} = 0[/tex]
[tex]6.67*10^{-11}*(\frac{4\pi}3{}2000*r^2)=\frac{1}{2}*2.8^2[/tex]
[tex]r = \sqrt{\frac{1}{2}*\frac{2.8^2}{6.67*10^{-11}*2000*4/3*\pi}}[/tex]
[tex]r=2648 m[/tex]
Therefore the diameter is 5296 m.
b) Applying the concept of gravitational force and centripetal force we have to
[tex]F_g = F_c[/tex]
[tex]\frac{G M m}{r^2} = \frac{m v^2}{r}[/tex]
[tex]\frac{G M}{r} = v^2[/tex]
[tex]6.67*10^{-11}*\frac{(\frac{4\pi}{3}2000*r^3)}{r} = 2.8^2[/tex]
[tex]r=3746 m[/tex]
Therefore the diameter would be 7492 m
The diameter of the largest asteroid from which a froghopper could jump free can be calculated using the escape velocity formula. Similarly, for the froghopper to enter a circular orbit, its horizontal speed needs to equal the required orbital speed, calculated using the gravitational field strength at the surface of the asteroid.
Explanation:To find the diameter of the largest spherical asteroid from which a froghopper could jump free, we need to understand the escape speed, the speed necessary to overcome an object's gravitational field. The formula for escape velocity is √(2*G*M/R), where G is the gravitational constant, M is the mass of the asteroid, and R is its radius (half of the diameter).
In this case, we can solve the formula for the radius, R = 2*G*M/v², where v is the takeoff speed of a froghopper (2.8 m/s). The mass of the spherical asteroid can be calculated from the given density (2.0 g/cm³ or 2000 kg/m³) and the volume of a sphere (4/3*π*R³). Substituting the mass into the formula will provide the radius, and hence diameter, of the asteroid.
For a froghopper to enter a circular orbit just above the surface of the asteroid, its horizontal speed must equal the circular orbital speed, which depends on the gravitational field strength at the surface of the asteroid. It's determined by the formula v = √(G*M/R), which can be rearranged to solve for the required diameter of the asteroid.
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How long must a simple pendulum be if it is to make exactly one swing per four seconds? (That is, one complete oscillation takes exactly 8.0 s.) Express your answer to two significant figures and include the appropriate units.
Answer:
Length of the pendulum will be 3.987 m
Explanation:
We have given time period of the pendulum T = 8 sec
Acceleration due to gravity [tex]g=9.81m/sec^2[/tex]
We have to find the length of the simple pendulum
We know that time period of the simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]8=2\times 3.14 \sqrt{\frac{l}{9.81}}[/tex]
[tex]l=3.987m[/tex]
So length of the pendulum will be 3.987 m
You drive 4 km at 30 km/h and then another 4 km at 50 km/h. What is your average speed for the whole 8-km trip?
a. Less than 40 km/h.
b. Not enough information.
c. Equal to 40 km/h.
d. More than 40 km/h.
Answer:
option A
Explanation:
given,
drive 4 km at 30 km/h and
another 4 km at 50 km/h
average speed for the whole 8 Km trip = ?
time for the trip of first 4 km
distance = speed x time
[tex]t_1 = \dfrac{d}{s}[/tex]
[tex]t_1 = \dfrac{4}{30}[/tex]
t₁= 0.1333 hr
time from another 4 km trip
[tex]t_2 = \dfrac{d}{s}[/tex]
[tex]t_2= \dfrac{4}{50}[/tex]
t₂ = 0.08 hr
average speed
[tex]s= \dfrac{d}{t}[/tex]
[tex]s= \dfrac{8}{0.1333 + 0.08}[/tex]
[tex]s= \dfrac{8}{0.2133}[/tex]
s = 37.50 Km/hr
Less than 40 Km/h
the correct answer is option A
A 3.49 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 111 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 10.3 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
Answer:
18.2 cm
Explanation:
The kinetic energy of the block
= .5 x 3.49 x 10⁻² x 10.3²
= 1.85 J
If x be the compression created in the spring due to this kinetic energy
1.85 = 1/2 k x²
= .5 x 111 x²
x² = 3.3 x 10⁻²
x = 0.182 m .
= 18.2 cm
This will be the amplitude of oscillation under SHM.
When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of 5.70 10-7 m and a maximum speed of 2.66 10-3 m/s. (a) What is the frequency (in Hz) of the eardrum's vibration
Answer:
Frequency will be 743.09 Hz
Explanation:
We have given maximum speed [tex]v_m=2.66\times 10^{-3}m/sec[/tex]
Amplitude [tex]A=5.7\times 10^{-7}m[/tex]
We have to find the frequency of the vibration
We know that angular frequency is given by
[tex]\omega =\frac{v_m}{A}=\frac{2.66\times 10^{-3}}{5.7\times 10^{-7}}=4666.666rad/sec[/tex]
[tex]2\pi f=4666.66[/tex]
[tex]f=743.09Hz[/tex]
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022 kg and a radius of 1821 km . For this calculation, ignore any variation in gravity over the 500-km range of the debris.
Final answer:
Io remains volcanically active due to tidal heating caused by Jupiter's gravitational forces.
Explanation:
Io, the innermost of Jupiter's Galilean moons, is known for its high level of volcanism, making it the most volcanically active body in the solar system. Despite its small size, Io remains volcanically active due to the effect of gravity, specifically tidal heating caused by the massive gravity of Jupiter. Jupiter's strong gravitational forces pull Io into an elongated shape, resulting in tidal heating that generates enough internal heat to drive volcanic activity on the moon.
A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as heat to the surroundings.) The average wavelength of the visible light is about 600 nm, so make the simplifying assumption that all the light has this wavelength. How many visible-light photons does the bulb emit per second?
The incandescent lightbulb emits approximately [tex]\(1.51 \times 10^{19}\)[/tex] visible-light photons per second.
To find the number of photons emitted per second, you can use the following steps:
1. Calculate the energy of one photon using the formula:
[tex]\[ E = hf \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of one photon,
- [tex]\( h \)[/tex] is Planck's constant [tex](\( 6.626 \times 10^{-34} \ \text{Joule}\cdot\text{s} \))[/tex],
- [tex]\( f \)[/tex]is the frequency of the light.
The frequency [tex](\( f \))[/tex] can be determined using the speed of light [tex](\( c \))[/tex] and the wavelength [tex](\( \lambda \))[/tex]:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
where:
- [tex]\( c \)[/tex] is the speed of light [tex](\( 3.00 \times 10^8 \ \text{m/s} \))[/tex],
- [tex]\( \lambda \)[/tex] is the wavelength of the light.
2. Find the number of photons emitted per second using the power of the visible light [tex](\( P_{\text{visible}} \))[/tex] and the energy of one photon:
[tex]\[ \text{Number of photons per second} = \frac{P_{\text{visible}}}{E} \][/tex]
where:
- [tex]\( P_{\text{visible}} \)[/tex] is the power of the visible light (5 W),
- [tex]\( E \)[/tex] is the energy of one photon.
Let's perform the calculations:
[tex]\[ f = \frac{c}{\lambda} \][/tex]
[tex]\[ f = \frac{3.00 \times 10^8 \ \text{m/s}}{600 \times 10^{-9} \ \text{m}} \][/tex]
[tex]\[ f = 5.00 \times 10^{14} \ \text{Hz} \][/tex]
Now, calculate the energy of one photon:
[tex]\[ E = hf \][/tex]
[tex]\[ E = (6.626 \times 10^{-34} \ \text{Joule}\cdot\text{s}) \times (5.00 \times 10^{14} \ \text{Hz}) \][/tex]
[tex]\[ E = 3.313 \times 10^{-19} \ \text{Joule} \][/tex]
Finally, calculate the number of photons emitted per second:
[tex]\[ \text{Number of photons per second} = \frac{5 \ \text{W}}{3.313 \times 10^{-19} \ \text{Joule/photon}} \][/tex]
[tex]\[ \text{Number of photons per second} \approx 1.51 \times 10^{19} \ \text{photons/s} \][/tex]
So, the incandescent lightbulb emits approximately [tex]\(1.51 \times 10^{19}\)[/tex] visible-light photons per second.
Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected in series with a 20e-2 F capacitor, a 20-mH inductor, and a 50- resistor. -1.37 radians 0.134 radians 0.0180 radians 0.0300 radians
Answer:
The phase angle is 0.0180 rad.
(c) is correct option.
Explanation:
Given that,
Voltage = 12 V
Angular velocity = 50 Hz
Capacitance [tex]C= 20\times10^{-2}\ F[/tex]
Inductance [tex]L=20\times10^{-3}\ H[/tex]
Resistance [tex]R= 50\ Omega[/tex]
We need to calculate the impedance
Using formula of impedance
[tex]z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}[/tex]
[tex]z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}[/tex]
[tex]z=50.00[/tex]
We need to calculate the phase angle
Using formula of phase angle
[tex]\theta=\cos^{-1}(\dfrac{R}{z})[/tex]
[tex]\theta=\cos^{-1}(\dfrac{50}{50.00})[/tex]
[tex]\theta=0.0180\ rad[/tex]
Hence, The phase angle is 0.0180 rad.
A student is doing a lab experiment in which she uses diffraction to measure the width of a shaft of her hair. She shines a 530 nm green laser pointer on a single hair, which produces a diffraction pattern on a screen 1.2 m away. The width of the central maximum of the pattern is 14 mm. a. What is the thickness of the hair? (91µm) b. If she chose a wider shaft of hair, how would this change the width of the central maximum? (b): For a single slit, we know that a wider slit gives a narrower pattern; the same is true for the hair.
Answer:
.09 mm
Explanation:
In case of diffraction by single slit with width a the width of central maxima is given below
width = 2 λD/a
where λ is wavelength of light , D is distance of screen , a is width o slit
substituting the given values
14 x 10⁻³ = [tex]\frac{2\times530\times10^{-9}\times1.2}{a}[/tex]
a = .09 mm
b) If a is greater , width of central maxima will be less wide.
A disc on a frictionless axle, starting from rest (0 rpm) can spin up to a rotation rate of 3820 rpm in a period of 2 seconds. (This is equivalent to angular acceleration of 200 rad/s.) Moment of inertia of disc = 5 kg-m. a) How much net torque was applied to the disc during the 2-s period? Answer: b) How much net torque would be needed to change the angular acceleration to 400 rad/s"? Answer: c) If the angular acceleration is 400 rad/s, how long will it take to spin up from 0 to 3820 rpm? Answer:
Answer:
1000 Nm
2000 Nm
1.00007 seconds
Explanation:
I = Moment of inertia = 5 kgm²
[tex]\alpha[/tex] = Angular acceleration
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
t = Time taken
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=5\times 200\\\Rightarrow \tau=1000\ Nm[/tex]
The torque of the disc would be 1000 Nm
If [tex]\alpha=400\ rad/s^2[/tex]
[tex]\tau=I\alpha\\\Rightarrow \tau=5\times 400\\\Rightarrow \tau=2000\ Nm[/tex]
The torque of the disc would be 2000 Nm
From equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{3820\times \frac{2\pi}{60}-0}{400}\\\Rightarrow t=1.00007\ s[/tex]
It would take 1.00007 seconds to reach 3820 rpm
A major artery with a 1.6 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area of 0.7 cm2. By what factor is the average velocity of the blood reduced when it passes into these branches?
To solve this problem it is necessary to apply the continuity equations for which it is defined that the proportion of Area in the initial section is equal to the final section. In other words,
[tex]A_1 v_1 = A_2 v_2[/tex]
Where,
[tex]A_i =[/tex] Cross sectional area at each section
[tex]v_i =[/tex]Velocities of fluid at each section
The total area of the branch is eighteen times of area of smaller artery. The average cross-sectional area of each artery is [tex]0.7cm^2.[/tex]
Therefore the Cross-sectional area at the end is
[tex]A_2= 18*0.7cm^2[/tex]
[tex]A_2 = 12.6cm^2[/tex]
Applying the previous equation we have then
[tex]A_1 v_1 = A_2 v_2[/tex]
[tex](1.7cm^2) v_1 = (12.6cm^2)v_2[/tex]
The ratio of the velocities then is
[tex]\frac{v_1}{v_2} = \frac{1.7}{12.6}[/tex]
[tex]\frac{v_1}{v_2} = 0.135[/tex]
Therefore the factor by which the velocity of blood will reduce when it enters the smaller arteries is 0.1349
A small, 0.10 kg cart is moving at 1.54 m/s when it collides with a larger 1.52 kg cart at rest. After the elastic collision, the small cart recoils (i.e. bounces off the larger cart and travels in the opposite direction) at a speed of 0.76 m/s.(a) What is the magnitude (absolute value) of the change in momentum for the small cart? kg m/s(b) What is the speed (absolute value of the velocity) of the larger cart after the collision? m/s
Answer:
a. the absolute value of the change in the momentum of the small car is 0.078
b. the velocity of the larger car after the collision is 0.1513 m/s
Explanation:
The linear momentum P is calculated as:
P = MV
Where M is the mass and V the velocity
Therefore, for calculated the change of the linear momentum of the small cart, we get:
[tex]P_{fc}-P_{ic} =[/tex]ΔP
where [tex]P_{ic}[/tex] in the inicial momentum and [tex]P_{fc}[/tex] is the final momentum of the small cart. Replacing the values, we get:
0.10 kg (0.76) -0.10(1.54) = -0.078 kg m/s
The absolute value: 0.078 kg m/s
On the other hand, using the law of the conservation of linear momentum, we get:
[tex]P_i = P_f[/tex]
Where [tex]P_i[/tex] is the linear momentum of the sistem before the collision and [tex]P_f[/tex] is the linear momentum after the collision.
[tex]P_i=P_{ic}\\P_f=P_{fc} + P_{ft}[/tex]
Where [tex]P_{fc}[/tex] is the linear momentum of the small cart after the collision and [tex]P_{ft}[/tex] is the linear momentum of the larger cart after the collision
so:
(0.10 kg)(1.54 m/s) = (0.10 kg)(-0.76 m/s) + (1.52 kg)(V)
Note: we choose the first direction of the small car as positive.
Solving for V:
[tex]\frac{0.10(1.54)+0.10(0.76)}{1.52} = V[/tex]
V = 0.1513 m/s
Final answer:
The magnitude of the change in momentum for the small cart is 0.078 kg·m/s. The speed of the larger cart after the collision is 0 m/s.
Explanation:
(a) What is the magnitude (absolute value) of the change in momentum for the small cart?
To find the change in momentum for the small cart, we can use the formula:
Change in momentum = mass × change in velocity
The mass of the small cart is 0.10 kg and its initial velocity is 1.54 m/s. After the collision, its velocity changes to -0.76 m/s (since it recoils in the opposite direction). Therefore, the change in velocity is: (0.76 m/s - 1.54 m/s) = -0.78 m/s.
Substituting the values into the formula, we get:
Change in momentum = 0.10 kg × (-0.78 m/s) = -0.078 kg·m/s
Since the question asks for the magnitude (absolute value) of the change in momentum, the answer is 0.078 kg·m/s.
(b) What is the speed (absolute value of the velocity) of the larger cart after the collision?
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the larger cart is initially at rest, its momentum before the collision is zero. Therefore, its momentum after the collision must also be zero.
Since momentum is equal to mass × velocity, the mass of the larger cart is 1.52 kg and its final velocity after the collision is denoted as Vf. We can set up the equation:
1.52 kg × Vf = 0 kg·m/s
Solving for Vf, we find that the velocity of the larger cart after the collision is also zero. Therefore, the answer is 0 m/s.
. In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring’s force constant
The spring's force constant is [tex]\( 15.30 \, \text{N/m} \).[/tex]
To find the spring's force constant, we can use the formula for the period of oscillation of a mass-spring system:
[tex]\[ T = 2\pi \sqrt{\frac{m}{k}} \][/tex]
where [tex]\( T \)[/tex] is the period of oscillation, m is the mass attached to the spring, and [tex]\( k \)[/tex] is the spring constant.
Given:
[tex]\[ m = 0.200 \, \text{kg} \]\[ \text{Time for one complete oscillation } = 2.60 \, \text{s} \][/tex]
First, let's find the period [tex]\( T \)[/tex] using the given time for one complete oscillation:
[tex]\[ T = 2.60 \, \text{s} \][/tex]
Now, we can rearrange the formula for the period to solve for the spring constant [tex]\( k \):[/tex]
[tex]\[ k = \frac{4\pi^2 m}{T^2} \][/tex]
Substitute the known values:
[tex]\[ k = \frac{4\pi^2 \cdot 0.200 \, \text{kg}}{(2.60 \, \text{s})^2} \][/tex]
Calculate:
[tex]\[ k = \frac{4\pi^2 \cdot 0.200 \, \text{kg}}{6.76 \, \text{s}^2} \]\[ k = \frac{7.854 \, \text{kg} \cdot \text{m}^{-1}}{6.76 \, \text{s}^2} \]\[ k \approx 15.30 \, \text{N/m} \][/tex]
So, the spring's force constant is [tex]\( 15.30 \, \text{N/m} \).[/tex]
Complete Question:
In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring’s force constant
A 155 kg satellite is orbiting on a circular orbit 5995 km above the Earth's surface. Determine the speed of the satellite. (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.)
Answer:[tex]5.67\times 10^{3} m/s[/tex]
Explanation:
Given
mass of satellite [tex]m=155 kg[/tex]
Satellite is orbiting 5995 km above Earth surface
mass of Earth [tex]M=5.97\times 10^{24} kg[/tex]
Radius of Earth [tex]R=6370 km[/tex]
here [tex]r=R+5995=6370+5995=12,365 km[/tex]
Gravitational Force will Provide the centripetal Force
[tex]\frac{GMm}{r^2}=\frac{mv^2}{r}[/tex]
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
[tex]v=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{12365\times 10^{3}}}[/tex]
[tex]v=\sqrt{32.203\times 10^6}[/tex]
[tex]v=5.67\times 10^3m/s=5674.7 km/s[/tex]
Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission is in the infrared region of the spectrum. The other has a temperature of 12,000 K, so its frequency of maximum emission is in the ultraviolet region of the spectrum.
Compared to the object at 3000 K, the object at 12,000 K:
A. emits more light in the infrared region of the spectrum.
B. emits more light in the visible region of the spectrum.
C. emits more light in the ultraviolet region of the spectrum
D. emits more light in only two of the three listed spectral regions.
E. emits more light in all three spectral regions.
The object at 12,000 K emits more light in the ultraviolet region of the spectrum compared to the object at 3000 K.
Explanation:When comparing the two objects, we can see that the object at 12,000 K emits more light in the ultraviolet region of the spectrum compared to the object at 3000 K. This is because as the temperature of an object increases, the frequency of maximum emission also increases. The object at 12,000 K has a higher frequency of maximum emission, which falls in the ultraviolet region of the spectrum.
A Garden hose is attached to a water faucet on one end and a spray nozzle on the other end. The water faucet is turned on, but the nozzle is turned off so that no water flows through the hose. The hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of .68m. What is the pressure inside the hose?
Answer:
The gauge pressure is 6670.8 Pascal
Explanation:
Given the height is 0.68 m
The Gauge pressure is determine by the formula,
Gauge pressure , P = hdg
Where h = 0.68 m
d = 1000
g = 9.81
Substituting the values in the formula of Gauge pressure, we get
Gauge Pressure = hdg =[tex]0.68 \times 1000 \times 9.81[/tex]
Gauge Pressure = 6670.8
Therefore, the gauge pressure is 6670.8 Pascal
The ratio of carbon-14 to nitrogen in an artifact is 1:1 . Given that the half-life of carbon-14 is 5730 years, how old is the artifact?
Answer: 5,728.5 years
Explanation:
By measuring the current ratio of daughter to parent, that is (Dt/Pt) one can deduce the age of any sample. The assumption is that there is no daughter atom present at time,t=0 and that the daughter atoms are due to the parent decay(where none has been lost).
Lamda= ln 2÷ half life,t(1/2)
Lamda= ln2= 0.693/5730 years
Lamda=1.21 × 10^-4
Using the formula below;
Age,t= 1/lamda× (ln {1+ Dt/Pt})------------------------------------------------------------------------------------(1)
Slotting our values into equation (1) above.
Age,t= 1/lamda(ln[1+1/1])
Age,t= 1/1.2×10^-4(ln 1+1)
Age,t= 1/1.21×10^-4(ln 2)
Age,t= ln 2/ 1.21×10^-4
Age, t= 5,728.5 years.
Answer:
11,460 years
Explanation:
half-life= 5,730 years
you are looking for the 1 from the ration 1:3
multiply the half-life by two in order to get the 1
5,730 x 2 = 11,640
=11,460 years
Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55.9 mph). The cars lock together in such a way as to prevent them from separating or rotating significantly. One second before the collision, the car is located at position (x,y) = (-20,0) m and the truck is located at position (x,y) = (0,-25) m.
1. Calculate the x-coordinate of the center of mass of the two-automobile system, xCM, one second before the collision. ________ m
2. Calculate the y-coordinate of the center of mass of the two-automobile system, xCM, one second before the collision.________ m
3. Calculate the magnitude of the velocity of the center of mass of the two-automobile system, vCM, one second before the collision. ________ m/s
4. Calculate the direction of the velocity of the center of mass of the two-automobile system, vCM one second before the collision. (Give you answer in degrees relative to the positive x-axis which points to the east.)
________ degrees
5. Calculate the magnitude of the velocity of the center of mass of the two-automobile system, vCM, immediately after the collision. ________ m/s
6. Calculate the direction of the velocity of the center of mass of the two-automobile system, vCM immediately after the collision. (Give you answer in degrees relative to the positive x-axis which points to the east.)
________ degrees
Answer:
1. [tex]X_{cm}=-8.57m[/tex]
2. [tex]Y_{cm}=-14.29m[/tex]
3. [tex]V_{cm} = 16.66m/s[/tex]
4. α = 59.05°
5. [tex]V_{cm} = 16.66m/s[/tex]
6. α = 59.05°
Explanation:
The position of the center of mass 1s before the collision is:
[tex]X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}[/tex]
where
[tex]X_c=-20m[/tex]; [tex]m_c=1500kg[/tex];
[tex]X_t=0m[/tex]; [tex]m_t=2000kg[/tex];
Replacing these values:
[tex]X_{cm}=-8.57m[/tex]
[tex]Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}[/tex]
where
[tex]Y_c=0m[/tex]; [tex]m_c=1500kg[/tex];
[tex]Y_t=-25m[/tex]; [tex]m_t=2000kg[/tex];
Replacing these values:
[tex]Y_{cm}=-14.29m[/tex]
The velocity of their center of mass is:
[tex]V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}[/tex]
where
[tex]V_{c-x}=20m/s[/tex]; [tex]m_c=1500kg[/tex];
[tex]V_{t-x}=0m/s[/tex]; [tex]m_t=2000kg[/tex];
Replacing these values:
[tex]V_{cm-x}=8.57m/s[/tex]
[tex]V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}[/tex]
where
[tex]V_{c-y}=0m[/tex]; [tex]m_c=1500kg[/tex];
[tex]V_{t-y}=-25m[/tex]; [tex]m_t=2000kg[/tex];
Replacing these values:
[tex]V_{cm-y}=-14.29m[/tex]
So, the magnitude of the velocity is:
[tex]V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}[/tex]
[tex]V_{cm}=16.66m/s[/tex]
The angle of the velocity is:
[tex]\alpha =atan(V_{cm-y}/V_{cm-x})[/tex]
[tex]\alpha=59.05\°[/tex]
Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east
Calculations using the physics concept of center of mass determine the positions and velocities of the system of two vehicles before and after collision. The x and y coordinates of the center of mass are found to be -12m and -15m respectively, with a combined vCM direction of 180 degrees relative to the positive x-axis.
Explanation:To answer the questions presented, it's important to recall the concept of center of mass in physics. The center of mass of a system of particles (like two cars in this case) is the point at which its mass may be considered to be concentrated, when calculating moments and linear momentum. The positions of center of mass are given by:
xCm = (m1*x1 + m2*x2) / (m1 + m2)
yCm = (m1*y1 + m2*y2) / (m1 + m2)
And the center of mass velocity can be found using:
vCm = (m1*v1 + m2*v2) / (m1 + m2)
Assuming east as positive x-direction and north as positive y-direction, substituting the given values into the equations above, we find that:
xCM is (-20 * 1500 + 0 * 2000) / (1500 + 2000) = -12 myCM is (0 * 1500 + -25 * 2000) / (1500 + 2000) = -15 mvCM is (20 * 1500 + 0 * 2000) / (1500 + 2000) = 12 m/sAnd since all motion is along the x and y axes, we can find the direction by calculating the arctangent of the ratio of the y-velocity to the x-velocity, yielding a vCM direction relative to the positive x-axis of 0 degrees. However, because the velocities are negative, the directions should be 180 degrees for xCM and 270 degrees for yCM. The combined direction would also be 180 degrees.Since the collision does not cause the vehicles to separate or rotate, vCM won't change and remains 12 m/s after the collision.Similarly, the direction of vCM won't change because the two cars are locked together.Learn more about Center of Mass here:https://brainly.com/question/33607198
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A major difference between AC and DC electricity is in its usefulness to effectively supply electricity to large groups. The best explanation for the difference is which of these statements?
A) DC current can be used in transformers while AC current cannot.
B) AC current can be used in transformers while DC current cannot.
C) AC current fluctuates too much to be used effectively in power generation.
D) DC current fluctuates too much to be used effectively in power generation.
Answer:
B) AC current can be used in transformers while DC current cannot.
Explanation:
AC current is used for long distance transportation of electrical energy with amplifying the voltage and reducing the current in the conductors so that there is minimum loss of energy in the form of heat.
The relation between heat energy and electrical current is given by Joule's law as:
[tex]Q=i^2.R.t[/tex]
where:
Q = amount of heat energy generatedi = current in the conductorR = resistance of the conductort = time for which the current flowsThe above mentioned variation of current and voltage in the transmission lines of AC is achieved by a device called transformer. It consist of a rectangular shaped steel core common to two of the insulated wire winding of which one side acts as input (primary coil) and the other one acts as secondary (output).
AC i.e. alternating current can only be used in the transformers because they have the continuously varying amplitude which in turn provides the change in flux with respect to the time inducing an emf in the nearby secondary coil of the transformer without any physically moving parts.
The alternating nature of AC current aloows changes in magnet fields to be used to transform electricity and make it more efficient to transport over long distances. AC current can be used in transfromers while DC current cannot.
The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling friction, the cart moves to the right with a constant velocity of 2 m/s. The discharge from the 50-mm-diameter nozzle is 0.04 m3/s. One-fourth of the discharge flows down the incline, and three-fourths flows up the incline. Assume steady flow, all within the vertical plane.
Answer
given,
constant speed of cart on right side = 2 m/s
diameter of nozzle = 50 mm = 0.05 m
discharge flow through nozzle = 0.04 m³
One-fourth of the discharge flows down the incline
three-fourths flows up the incline
Power = ?
Normal force i.e. Fn acting on the cart
[tex]F_n = \rho A (v - u)^2 sin \theta[/tex]
v is the velocity of jet
Q = A V
[tex]v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}[/tex]
[tex]v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}[/tex]
v = 20.37 m/s
u be the speed of cart assuming it to be u = 2 m/s
angle angle of inclination be 60°
now,
[tex]F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0[/tex]
F n = 2295 N
now force along x direction
[tex]F_x = F_n sin 60^0[/tex]
[tex]F_x = 2295 \times sin 60^0[/tex]
[tex]F_x = 1987.52\ N[/tex]
Power of the cart
P = F x v
P = 1987.52 x 20.37
P = 40485 watt
P = 40.5 kW
The power produced by the stream is; 40.5 kW
What is the power produced?We are given;
Constant speed of cart on right side; u = 2 m/s
Diameter of nozzle; d = 50 mm = 0.05 m
Discharge flow through nozzle; Q = 0.04 m³/s
Formula for the normal force acting on the cart is;
Fₙ = ρA(v - u)²sin θ
Let us find v the speed of the jet from;
Q = Av
v = Q/A
v = 0.04/(π * 0.05²/4)
v = 20.37 m/s
From online research about this question, the angle of inclination is 60°. Thus;
Fₙ = ρA(v - u)²sin θ
Fₙ = 1000 * (π * 0.05²/4) * (20.37 - 2)sin 60
Fₙ = 2295 N
Force along the x-direction will be;
Fₓ = Fₙ * sin 60
Fₓ = 2295 * sin 60
Fₓ = 1987.52 N
Power of the Cart is;
P = Fₓ * v
P = 1987.52 * 20.37
P = 40.5 kW
Read more about power at; https://brainly.com/question/25864308
A wire carries a current of 4.1 A. How many electrons per second are passing any cross sectional area of the wire? Enter your answer in the format *.**E** (for example, the fundamental unit of charge, which is 1.60 cross 10^(-19) C, would be entered by typing 1.60E-19). electrons per second
To solve this problem it is necessary to apply the concepts related to Current and Load.
The current in terms of the charge of an electron can be expressed as
[tex]i = \frac{q}{t}[/tex]
Where,
q = Charge
t = time
At the same time the Charge is the amount of electrons multiplied by the amount of these, that is
q = ne
Replacing in the first equation we have to
[tex]i = \frac{q}{t}[/tex]
[tex]i = \frac{ne}{t}[/tex]
Clearing n,
[tex]n = \frac{it}{e}[/tex]
Here the time is one second then
[tex]n = \frac{i}{e}[/tex]
[tex]n = \frac{4.1}{1.6*10^{-9}}[/tex]
[tex]n = 2.56*10^{19}electrons[/tex]
Therefore the number of electrons per second are passing any cross sectional area of the wire are [tex]2.56*10^{19}electrons[/tex]
|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and reducing the force on the climber. In one standardized test of ropes, an 80 kg mass falls 4.8 m before being caught by a 2.5-m-long rope. If the net force on the mass must be kept below 11 kN, what is the minimum time for the mass to come to rest at the end of the fall
To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
[tex]F = ma[/tex]
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{11*10^3}{80}[/tex]
[tex]a = 137.5m/s[/tex]
From the cinematic equations of motion we know that
[tex]v_f^2-v_i^2 = 2ax[/tex]
Where,
[tex]v_f =[/tex]Final velocity
[tex]v_i =[/tex]Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
[tex]v_f^2-v_i^2 = 2ax[/tex]
[tex]0-v_i^2 = 2(-g)x[/tex]
[tex]v_i =\sqrt{2gx}[/tex]
[tex]v_i = \sqrt{2*9.8*4.8}[/tex]
[tex]v_i = 9.69m/s[/tex]
From the equation of motion where acceleration is equal to the velocity in function of time we have
[tex]a = \frac{v_i}{t}[/tex]
[tex]t = \frac{v_i}{a}[/tex]
[tex]t =\frac{9.69}{137.5}[/tex]
[tex]t = 0.0705s[/tex]
Therefore the time required is 0.0705s
The minimum time for the mass to come to rest at the end of the fall is about 0.071 s
[tex]\texttt{ }[/tex]
Further explanationNewton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
[tex]\texttt{ }[/tex]
[tex]\large {\boxed {F = \Delta (mv) \div t }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
v = Velocity of Object ( m/s )
t = Time Taken ( s )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of climber = m = 80 kg
height of fall = h = 4.8 m
net force = ∑F = 11 k N = 11000 N
Asked:
minimum time = t = ?
Solution:
FIrstly , we could find the initial velocity of climber as he caught by the rope:
[tex]v^2 = u^2 + 2gh[/tex]
[tex]v^2 = 0^2 + 2(9.8)(4.8)[/tex]
[tex]v^2 = 94.08[/tex]
[tex]v = \frac{28}{3}\sqrt{5} \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
Next , we will use Newton's Law of Motion to calculate the minimum time:
[tex]\Sigma F = \Delta p \div t[/tex]
[tex]\Sigma F = m \Delta v \div t[/tex]
[tex]11000 = 80 (\frac{28}{3}\sqrt{5}) \div t[/tex]
[tex]t = 80 (\frac{28}{3}\sqrt{5}) \div 11000[/tex]
[tex]t \approx 0.071 \texttt{ s}[/tex]
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441Newton's Law of Motion: https://brainly.com/question/10431582Example of Newton's Law: https://brainly.com/question/498822[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Dynamics
A 145.0 km long electric power line made out of copper is installed during the Spring months at an average temperature of 13.7 °C. What is the change in the length of the power line on a hot Summer day, when the temperature is 42.0 °C? The coefficient of linear expansion for copper is 1.62×10-5 1/C°. Use positive sign for an increase in length and negative for a decrease.
Answer:
66.4767 m
Explanation:
[tex]L_0[/tex] = Original length of power line = 145 km
[tex]\alpha[/tex] = Coefficient of linear expansion = [tex]1.62\times 10^{-5}\ /^{\circ}C[/tex]
Initial temperature = 13.7°C
Final temperature = 42°C
Change in length of a material is given by
[tex]\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=1.62\times 10^{-5}\times 145000\times (42-13.7)\\\Rightarrow \Delta=66.4767\ m[/tex]
The change in length will be 66.4767 m