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lawyer [7]
2 years ago
11

Two rods are identical, except that one is brass (Y = 9.0 × 1010 N/m2) and one is tungsten (Y = 3.6 × 1011 N/m2). A force causes

the brass rod to stretch by 3.0 × 10-6 m. What is the amount of stretch ΔLTungsten for the tungsten rod when the same force is applied to it?
Physics
1 answer:
PIT_PIT [208]2 years ago
8 0

To solve this problem it is necessary to apply the concepts related to Young's Module, and find the radius that gives the ratio between the two given materials. Young's module can be defined as,

Y=\frac{FL}{A \Delta L}

Where,

F= Force

L = Initial Length

A = Cross-sectional Area

\Delta L = Change in Length

Re-arrange the equation to find the change in Length we have,

\Delta L = \frac{FL}{AY}

If both the Force, as the Area and the initial length are considered constant, we can realize directly that the change in length is inversely proportional to Young's Module, therefore

\Delta L \propto \frac{1}{Y}

Applying this concept to that of the two materials (Brass and Tungsten),

\frac{\Delta L_T}{\Delta L_B} = \frac{Y_B}{Y_T}

\frac{\Delta L_T}{\Delta L_B} = \frac{9*10^{10}}{3.6*10^{11}}

\frac{\Delta L_T}{\Delta L_B} = 0.25

If the force caused 3 * 10^ {- 6}m to be stretched, the tungsten will stretch 0.25 of that ratio

L_T = 3*10^{-6}*0.25

L_T = 7.5*10^{-7}m

Therefore the amount of stretch of Tungsten is 7.5*10^{-7}m

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
2 years ago
A footballer kicks a ball at an angle of 45° with the horizontal. If the ball was in the air
Naily [24]

Answer:

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3 0
2 years ago
Driving along a boring stretch of interstate in Illinois, you start experimenting using the average speed equation you learned i
astra-53 [7]
The average speed would be 33.29m/s.
The average speed equation is:

Average speed =  \frac{total distance}{total time}

First you will need to solve for the distance you traveled in each scenario. So we can solve this by getting the product of speed and the time traveled. 

Scenario 1:
Speed = 29m/s
Time = 120s
Distance = ?

Distance = (29m/s)(120s)
               = 3,480m

Scenario 2
Speed = 35m/s
Time = 300s
Distance = ? 

Distance = (35m/s)(300s)
               = 10,500m

Now that you have the distance of both, you can solve for your average speed. 

Average speed = \frac{total distance}{total time}
                                = \frac{3,480m+10,500m}{120s+300s}
                                = \frac{13,980m}{420s}
                                = 33.29m/s
5 0
3 years ago
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Arturiano [62]
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2 years ago
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