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4 years ago

How does weathering affect statues??

Physics

2 answers:

Mamont248 [21]4 years ago

6 0

It breaks down because weathering is the breaking down of rocks and minerals

Harman [31]4 years ago

5 0

The weathering can affect statues because if it never rains for a long time we will be miserable and low on food, sand we be flowing in the air which will be bad because if we breathe to much in we can get very sick. Another thing is if it gets to cold we can get sick and if it's way below freezing we can die. With heat we know someday it will get to hot and we will all die or the sun will burn out and we will die.

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8. What is the S.l unit for mass?<br> A. kg<br> B. m/s2<br> D.N<br> C. m/s

katrin2010 [14]

Answer:

the answer is kg

Explanation:

because the s.i unit for mass is kilogram

3 0

3 years ago

If the change in velocity increases, what happens to the acceleration during the same time period?

vfiekz [6]

The answer should be “Acceleration increases.”

4 0

3 years ago

Read 2 more answers

A 16.9 kg monkey is swinging on a

mrs_skeptik [129]

The potential energy of the monkey at the given position is 600.9 J.

<h3>What is potential energy?</h3>

Potential energy is the energy stored in a body that depends upon the relative position of various parts of the body.

It increases with increase in the height of the object above the ground level.

<h3>Potential energy of the monkey</h3>

The potential energy of the monkey is calculated as follows;

P.E = mgh

P.E = mg (L sin43)

where;

m is mass of the monkey, given as 16.9 kg
g is acceleration due to gravity, given as 9.8 m/s²
L is length of the vine given as 5.32 m

Substitute the given parameters and solve potential energy of the monkey

P.E = (16.9)(9.8)( 5.32 x sin 43)

P.E = 600.9 J

Thus, the potential energy of the monkey at the given position is 600.9 J.

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

5 0

2 years ago

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

Fiesta28 [93]

Answer:

a) $P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

b) For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

Explanation:

For this case we have the following data given:

$D= 1mm = 0.001 m$ represent the diameter

$r = D/2= 0.0005m$ represent the radius

$T= 5.7 N$ represent the tension

$f = 57 Hz$ represent the frequency of the oscillator

$A= 0.54 cm = 0.0054 m$ represent the amplitude of the wave

Part a

For this case we can assume that the power transmitted to the wave is the same power of the oscillator. and we have the following formula for the power:

$P= 2 \pi^2 \rho S v f^3 A^2$

This expression can be written in different ways:

$P= 2 \pi^2 \rho S \sqrt{\frac{T}{\mu}} f^2 A^2$

$P= 2\pi^2 \rho S \sqrt{\frac{T}{\rho S}} f^2 A^2$

$P= 2 \pi^2 f^2 A^2 \sqrt{S \rho T}$

Where f is the frequency , T the tension $rho= 7800 \frac{kg}{m^3}$ the density of the steel, A the amplitude and $S= \pi r^2$ the area, so then we have everuthing in order to replace and we got:

$P = 2 \pi^2 (57 Hz)^2 (0.0054 m)^2 \sqrt{\pi(0.0005m)^2 (7800 \frac{kg}{m^3}) (5.7N)}= 0.349 W$

Part b

For this case we se that $P \propto A^2 f^2$

Since the power is constant but the frequency is doubled we will see that $A^2 \propto \frac{P}{f^2}$

So the original amplitude is $A_i \propto \sqrt{\frac{P}{f^2}}$

And if the frequency is doubled we have that:

$A^2_f \propto \frac{P}{(2f)^2}= \frac{P}{4f^2}$

$A_f \propto \sqrt{\frac{P}{4f^2}}= \frac{1}{2} \sqrt{\frac{P}{f^2}}$

So then we will see that the amplitude would be reduced the half and for this case:

$A_f = \frac{0.54cm}{2}= 0.27 cm = 0.0027 m$

8 0

4 years ago

A string of length 1.3 m is oscillating in a standing wave pattern. If the tension in the string is 430 N, the string has a mass

Vlad1618 [11]

To solve this problem we will use the concepts related to the speed of a string which is given by the applied voltage and the linear mass density of it. With the speed value we can find the fundamental frequency that will serve as a step to find the maximum speed through the relation of Amplitude and Angular Speed. So:

$v = \sqrt{\frac{T}{\mu_e}}$