13.

Can you think of a scenario when the kinetic and gravitational potential energy could both be zero ? Describe or draw how this c

Inga [223]

Both kinetic and gravitational potential energy can become zero at infinite distance from the Earth.

Consider an object of mass m projected from the surface of the Earth with a velocity v.

The total energy of the body on the surface of the Earth is the sum of its kinetic energy $\frac{1}{2} mv^2$ and gravitational potential energy $-\frac{GMm}{R^2}$ .

here, M is the mass of the Earth, R is the radius of Earth and G is the universal gravitational constant.

The gravitational potential energy of the object is negative since it is in an attractive field, which is the gravitational field of the Earth.

The energy of the object on the surface of the earth is given by,

$E_i=\frac{1}{2} mv^2-\frac{GMm}{R^2}$

As the object rises upwards, it experiences deceleration due to the gravitational force of the Earth. Its velocity decreases and hence its kinetic energy decreases.

The decrease in kinetic energy is manifested as an equal increase in potential energy. The potential energy becomes less and less negative as more and more kinetic energy is converted into potential energy.

At a height h from the surface of the Earth, the energy of the object is given by,

$E_h=\frac{1}{2} mv_h^2-\frac{GMm}{(R+h)^2}$

The velocity $v_h$ is less than v.

When h =∞, the gravitational potential energy increases from a negative value to zero.

If the velocity of projection is adjusted in such a manner that the velocity decreases to zero at infinite distance from the earth, the object's kinetic energy also becomes equal to zero.

Thus, it is possible for both kinetic and potential energies to be zero at infinite distance from the Earth. In this case, kinetic energy decreases from a positive value to zero and the gravitational potential energy increases from a negative value to zero.

7 0

3 years ago

A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

$v = \sqrt{\frac{2GM}{R}}$

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

$v_o = \sqrt{\frac{GM}{R+h}}$

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

$T = \frac{2\pi(R+h)}{v_o}$

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

3 years ago

What kind of system is generally composed of a fluid system to move the heat from the collector to its point of usage and a rese

Mkey [24]

Solar heating is the system composed of a fluid system to move the heat from the collector to its point of usage and a reservoir to stock the heat

Explanation:

The options given here like coal burning uses solid material as the source to heat and to generate energy. Similarly, nuclear power also requires solid particles like atoms or neutrons to strike the moderators forming energy.

In both of these cases, fluid system is present but it is used completely as coolant and to maintain the temperature. Thus, the remaining system that is solar heating has been done for water tanks where the fluid as water is used to move the heat from its collector to its point of usage. Even in solar system it is used as reservoir to stock the heat.

5 0

3 years ago

A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving

Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

$m_av_a + m_sv_s = 0$

Where $m_a = 92kg$ is the mass of the astronaut, $m_s = 1200kg$ is the mass of the satellite, $v_s = 0.14 m/s$ is the speed of the satellite. We can calculate the speed $v_a$ of the astronaut:

$v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s$

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0

3 years ago

there are 1.6 km in a mile. the distance between two cities is 248 miles. How many kilometers apart are the two cities?

podryga [215]

1.6 X 248=399.117
hope this helps

3 0

3 years ago