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lisabon 2012 [21]
2 years ago
6

As your rockets went upwards how would you describe how the energies changed?

Physics
1 answer:
chubhunter [2.5K]2 years ago
4 0

As a rocket increases height and slows down, it gains more and more potential energy and loses more and more kinetic energy. Potential energy is store energy (usually determined by height), and kinetic energy increases as speed increases.

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State and briefly describe the applications of expansion<br>​
zlopas [31]
When solid material expands in response to an increase in temperature (thermal expansion), it can increase in length in a process known as linear expansion. for an example application of expansion and contraction.

examples =
(1) Changing of shape and dimensions of objects such as doors.
(2) Wall collapsing due to bulging.
(3) Cracking of glass tumbler due to heating.
(4) Bursting of metal pipes carrying hot water or steam are some of the disadvantages of thermal expansion of matter.
6 0
2 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?
pishuonlain [190]

Answer: 800

Explanation:

1/2 x m x v^2 = m x g x h

KE = 10 x 10 x 8

KE= 800

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3 years ago
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Artist 52 [7]

Answer:

option (B) is the correct option.

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