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MAXImum [283]
3 years ago
8

Which object has the most gravitational potential energy?

Physics
1 answer:
Kipish [7]3 years ago
3 0

Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.

Explanation:

Gravitational potential energy is the product of mass of object, height of object and gravitational field.

So, formula to calculate gravitational potential energy is as follows.

U = mgh

where,

m = mass of object

g = gravitational field = 9.81 m/s^{2}

h = height of object

(A) m = 5 kg and h = 2m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J    (1 J = kg m^{2}/s^{2})

(B) m = 8 kg and h = 2 m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J    (1 J = kg m^{2}/s^{2})

(C) m = 8 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J    (1 J = kg m^{2}/s^{2})

(D) m = 5 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J    (1 J = kg m^{2}/s^{2})

Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.

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Which of the following would effect the speed of a wave ? A) the matter it travels through
Artemon [7]

Answer:

Option D

The frequency

Explanation:

The speed of wave is depedant only on the wavelength and frequency of waves since it is given by s=fw where s is the speed, f is frequency and w is the wavelength. Since the options given has only one factor, that is frequency, hence option D is correct. In case we had wavelength could be among the options, both would be correct.

5 0
3 years ago
A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

4 0
4 years ago
A 45.2-kg person is on a barrel ride at an amusement park. She stands on a platform with her back to the barrel wall. The 3.74-m
elena-14-01-66 [18.8K]

Answer:

  • <u><em>1,230N</em></u>

Explanation:

<u>1. Name of the variables:</u>

   f:frequency\\\\ \omega:angular\text{ }speed\\\\ a_c:centripetal\text{ }acceleration\\\\ F_c:centripetal\text{ }force\\ \\ m:mass\\ \\ d:diameter\\ \\ r:radius\\ \\ g:gravitational\text{ }acceleration

<u>2. Formulae:</u>

         f=\dfrac{number\text{ }of\text{ }revolutions}{time}

          \omega=2\pi f

          a_c=\omega^2 r

           F_c=m\times a_c

<u>3. Solution (calculations)</u>

       f=\dfrac{1}{1.65s}=0.\overline{60}s^{-1}

       \omega=2\pi\times0.\overline{60}\approx 3.808rad/s

      a_c=(3.808rad/s)^2\times (3.74/2m)=27.12m/s^2

      F_c=45.2kg\times27.12m/s^2=1,225.67N\approx 1,230N

3 0
3 years ago
Imagine tying a string to a ball and twirling it around you. How is this similar to the moon orbiting Earth? In this example, wh
leonid [27]
The string kind of acts like gravity 
5 0
4 years ago
This when a satellite orbits in an oval-shaped path around a central object.
Ilya [14]

Elliptical orbit.<<<<<<<<<<



8 0
3 years ago
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