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3 years ago

12

An object that is in free fall seems to be

1 answer:

Ratling [72]3 years ago

5 0

<span>An object that is in free fall seems to be weightless, even though the
usual force of gravity is acting on it.

If the falling object happens to be a person, then every part of him is
falling at the same rate, so none of his internal organs are pressing
against anything like they normally do, and he has no sensation of weight.

If a bathroom scale happens to be falling with him and he places it under
his feet, it won't read anything, since the scale and the person are both
falling at the same rate. </span>

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Need help with this worksheet. Very short

attashe74 [19]

Answer:

In the grapher: The stops are marked with a flat line, velocity with a diagonal line, and acceleration with a curve.

Average speed= Total distance/Total time

Explanation:

6 0

2 years ago

What is the entropy of a closed system in which 28 distinguishable grains of sand are distributed among 1000 distinguishable equ

baherus [9]

Answer:

193.4 J/K

Explanation:

For a closed system, the entropy is given as the natural logarithm of microstates.

The number of particles ( grains of sand ) N = 28

The number of boxes ( compartments ) M = 1000

Entropy is the logarithm of number of microstates

$S=ln \psi$

But the number of microstates is given by

$\psi=M^{N} hence S=ln M^{N}=\ln (1000^{28})= \boxed{193.4 J/K}$

4 0

2 years ago

A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

$V = \sqrt{\frac{T}{\mu}}$

Our values are given as:

$f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w$

Replacing we have that the velocity is

$V = \sqrt{\frac{T}{\mu}}$

$V = \sqrt{\frac{65}{0.230}}$

$V = 16.81m/s$

From the theory of wave propagation the average power wave is given as

$P =\frac{1}{2} \mu \omega^2 A^2 V$

Where,

A = Amplitude

$\omega = 2\pi f \rightarrow$ Angular velocity

$A^2 = \frac{2P}{\mu \omega^2 V}$

$A^2 = \frac{2P}{\mu (2\pi f)^2 V}$

Replacing,

$A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}$

$A = 0.0165m$

Therefore the amplitude of the wave should be 0.0165m

8 0

3 years ago

1 point<br> How much force is needed to accelerate an 84-kg boulder at a rate of 6.4<br> m/s/s? *

Katen [24]

Answer:

<h2>537.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 84 × 6.4

We have the final answer as

<h3>537.6 N</h3>

Hope this helps you

5 0

2 years ago

Which list places the layers of the sun in the correct order from outermost to innermost?

Yuki888 [10]

Is D

is D because the inner layers are the Core, Radiative Zone and Convection Zone.

5 0

2 years ago