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3 years ago

Which of the following is not an example of approximate simple harmonic motion? A. a ball bouncing on the floor

Physics

1 answer:

Reptile [31]3 years ago

3 0

<h2>Answer:</h2>

Answer to this question is (A)

<h2>Explanation</h2>

A ball bouncing on the floor is not the example of simple harmonic motion. SHM is the special kind of to and fro motion in which a particle oscillate about its mean position in a straight line. The acceleration of the particle is always directed towards its mean position and is directly proportional to its displacement from its mean position.

In case of a ball bouncing on the ground, the motion of the ball is not SHM, as neither it’s a to and fro motion nor the acceleration is proportional to its displacement from its mean position.

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In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

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2 years ago

A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg

ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

$Area = A=2.54*10^-^3m^2\\Force = F = 1.01*10^4N\\density = p = 7860 kg/m^3$

Required:

Speed of Traverse wave = c =?

Solution:

As we know that

$p=m/V\\\\ p=m/(L*A)\\p*A=m/L$

Now the equation for speed of traverse wave is calculated through:

$\sqrt \frac{F*L}{m}\\$

= $\sqrt\frac{F}{m/L} \\\sqrt{} \frac{F}{p*A}$

Substituting the values

$\sqrt\frac{1.01*10^4}{7860*2.54*10^-^3} \\$

=22.49 m/s

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3 years ago

Your friend says that inertia is a force that keeps things in their place, either at rest or in motion. Do you and your discussi

Galina-37 [17]

Answer:

yes i agree

Explanation:

because law of inertia state that object remain at rest or in motion unless external force apply on it

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2 years ago

8. Differentiate between each of these land designations made during the Progressive Era:

aliya0001 [1]

Answer:djhssnjsksksmksjsbhddbd

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3 years ago

A spring with a constant k=400n/m shoots a 1. 00kg ball up a frictionless incline after being compressed 0. 150m. what is the ma

Zielflug [23.3K]

The maximum height reached by the ball is 0.46m.

To find the answer, we have to know about the potential energy of a spring mass system.

<h3>How to find the maximum height reached by the ball?</h3>

It is given that,

$k=400N/m\\m=1kg\\x=0.150m\\h=?\\$

We have to find the maximum height reached by the ball.
Thus, we have the expression for potential energy of a spring mass system and that of gravitational field as,

$U=\frac{1}{2}kx^2 \\U=mgh$

Combining both, we get,

$h=\frac{kx^2}{2mg} =\frac{400*(0.15)^2}{2*1*9.8} =0.46m$

Thus, we can conclude that, the maximum height reached by the ball is 0.46m.

Learn more about the potential energy here:

brainly.com/question/26962934

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7 months ago