Answer:
The magnitudes of the speed and acceleration of the Earth are approximately 66,661.217 miles per hour and 47.782 miles per square hour, respectively.
Explanation:
Given that the Earth has a circular orbit and make a revolution at constant speed around the Sun. Then, the kinematic formulas for the speed and acceleration of the planet are, respectively:
Speed
(1)
Acceleration
(2)
Where:
- Speed of the planet, measured in miles per hour.
- Acceleration of the planet, measured in miles per square hour.
- Radius of the orbit, measured in miles.
- Period of rotation, measured in hours.
If we know that 
and 
, then the magnitudes of the speed and acceleration of the planet is:
The magnitudes of the speed and acceleration of the Earth are approximately 66,661.217 miles per hour and 47.782 miles per square hour, respectively.
Potential and kinetic energy are at play when we talk about Newton's second law of motion through the various positions in relation to the bodies involved.
<h3>What is Newton's second law of motion?</h3>
This law states that force is equal to the rate of change of momentum and is denoted as F = mv where m is mass and v is velocity.
Potential energy is the energy is possessed by a body by virtue of its position while kinetic energy is possessed by a body by virtue of its motion. Both forms of energy are influenced by forces and are equal to the total momentum.
Read more about Newton's second law of motion here brainly.com/question/2009830
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Wood is a poor conductor and therefore a good insulator keeping heat inside
Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
