Answer: Speeding up the orbital speed of earth so it escapes the sun require the greater energy.
Explanation: To find the answer, we need to know more about the Orbital and escape velocities.
<h3>
What is Orbital and Escape velocity?</h3>
- Orbital velocity can be defined as the minimum velocity required to put the satellite in its orbit around the earth.
- The expression for orbital velocity near to the surface of earth will be,
- Escape velocity can be defined as the minimum velocity with which a body must be projected from the surface of earth, so that it escapes from the gravitational field of earth.
- The expression for orbital velocity will be,
- If we want to get into the sun, we want to slow down almost completely, so that your speed relative to the sun became almost zero.
- We need about twice the raw speed to go to the sun than to leave the sun.
Thus, we can conclude that, the speeding up the orbital speed of earth so it escapes the sun require the greater energy.
Learn more about orbital and escape velocity here:
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The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Given the data in the question;
Since the brick was initially at rest before it was dropped,
- Initial Velocity;
- Height from which it has dropped;
- Gravitational field strength;
Final speed of brick as it hits the ground; 
<h3>Velocity</h3>
velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:
Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.
To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.
Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Learn more about equations of motion: brainly.com/question/18486505
Explanation:
It varies with altitude, but at sea level, it's 9.8 m/s².
The work done on the puck is 96 J
Explanation:
According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.
Mathematically:
where
is the final kinetic energy of the puck, with
m = 2 kg being the mass of the puck
v = 10 m/s is the final speed
is the initial kinetic energy of the puck, with
u = 2 m/s being the initial speed of the puck
Substituting numbers into the equation, we find the work done by the player on the puck:
Learn more about work and kinetic energy:
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