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3 years ago

What happens to warm air when it cools?

Physics

2 answers:

MrMuchimi3 years ago

7 0

Answer:

I think the answer is B

Explanation:

The warm air turns cold and then it goes back to clouds

wolverine [178]3 years ago

4 0

Answer:

b I'm pretty sure sorry if I'm wrong

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A book is on the table. If the weight of the book is 25 newtons, what is the magnitude and direction of the normal force?

Andrej [43]

I'm not sure about the magnitude, but the direction of the normal force is upward.

4 0

3 years ago

Read 2 more answers

a spring has a spring of 120 newtons per meter. calculate the elastic potential energy stored in the spring when it is stretched

JulsSmile [24]

We know that force <span>for a linear spring is equal to F=kx. Since energy is the potential to do work, and the spring is assumed to be perfectly elastic. Work can be written as:

W = F (integral sign) d</span>x

Therefore,

<span>W=k(Δx<span>)^2 / 2
W = F </span></span>Δx / 2
<span>W = 120 (2.0) / 2
W = 120

</span>

6 0

3 years ago

The light from a nearby star is observed to be red-shifted by 10% Is the star approaching or receding from the earth? How fast i

Romashka-Z-Leto [24]

Answer:

3.149×10^7 m/s

Explanation:

the frequency is shifted to lower value hence the star is receding the Earth.

$f=f_0\sqrt{\frac{1-\beta}{1+\beta} }$ ................1

f= 0.9 f_0

putting the value in 1 we get

$0.9f_0=f_0\sqrt{\frac{1-\beta}{1+\beta} }$ .

$0.9=\sqrt{\frac{1-\beta}{1+\beta} }$ .

on solving we get

β= 0.1049

now

$\frac{v}{c} = \beta$

v= 0.1049×3×10^8= 3.149×10^7 m/s

5 0

3 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$