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MA_775_DIABLO [31]
3 years ago
8

A block whose weight is 45 N rests on a horizontal table. A horizontal force of 36 N is applied to the block. The coefficient of

kinetic and static friction are 0.65 and 0.42, respectively. Will the block move under the influence of the force, and if so, what will be the blocks acceleration?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

1.5 m/s²

Explanation:

For the block to move, it must first overcome the static friction.

Fs = N μs

Fs = (45 N) (0.42)

Fs = 18.9 N

This is less than the 36 N applied, so the block will move.  Since the block is moving, kinetic friction takes over.  To find the block's acceleration, use Newton's second law:

∑F = ma

F − N μk = ma

36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a

6.75 N = 4.59 kg a

a = 1.47 m/s²

Rounded to two significant figures, the block's acceleration is 1.5 m/s².

Usually the coefficient of static friction is greater than the coefficient of kinetic friction.  You might want to double check the problem statement, just to be sure.

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The absence of external forces will make the pucks move in the form of a uniform circular motion.

<h3>What is a circular motion?</h3>

It should be noted that a circular motion simply means the movement of an object along the circumference of the circle.

In this case, the absence of external forces will make the pucks move in the form of a uniform circular motion.

If the friction is absent, the pucks will continue to move on the same path due to the first law of Newton and the law of conversation of energy. In this case,the results will match the predictions until there's loss in energy.

Learn more about circular motion on:

brainly.com/question/106339

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2 years ago
I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an
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Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = \sqrt{\frac{T}{\frac{M}{L}}}

Frequency (F₁) = \frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = \frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0
3 years ago
What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An
GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}

E=\frac{2m_{0}c^}2}{\sqrt{3}}

E=\frac{2\times0.001\times9\times10^{16}}{1.732}

E=1.04\times10^{14}J

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV

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3 years ago
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Eddi Din [679]

Answer:

<h3>The answer is 40 N</h3>

Explanation:

The force acting on an object can be found by using the formula

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From the question

mass = 4 kg

acceleration = 10 m/s²

So we have

force = 4 × 10

We have the final answer as

<h3>40 N</h3>

Hope this helps you

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Answer:

Detailed solution is given below

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