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Nataly_w [17]
3 years ago
10

A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose

mass is about 1/9 and radius 1/2 that of Earth, is mostnearly
A) 1/3 T
B) 2/3T
C)T
D) 3/2T
E) 3 T
Physics
1 answer:
olchik [2.2K]3 years ago
6 0

Answer:

C)T

Explanation:

The period of a mass-spring system is:

T=2\pi\sqrt\frac{m}{k}

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

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sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

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a is the distance between the slits

In this problem,

n = 5

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a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

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y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

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Answer:

Quick Answer:

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Explanation:

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