An object is lifted from the surface of a spherical planet to an altitude equal to the radius of the planet.
As a result, the object's <em>mass remains the same</em>, and its <em>weight decreases</em> to 1/4 of whatever it is when the object is on the planet's surface.
The kinetic energy of the phone right before it hits the ground is 9J.
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Kinetic energy of the phone</h3>
The kinetic energy of the phone right before it hits the ground is calculated as follows;
K.E = ¹/₂mv²
where;
- m is mass of the phone
- v is velocity of the phone
K.E = ¹/₂(0.08)(15)²
K.E = 9 J
Thus, the kinetic energy of the phone right before it hits the ground is 9J.
Learn more about kinetic energy here: brainly.com/question/25959744
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Answer:
distance traveled is a total length of the path traveled between two positions.
Answer:
14 m/s²
Explanation:
Start with Newton's 2nd law: Fnet=ma, with F being force, m being mass, and a being acceleration. The applied forces on the left and right side of the block are equivalent, so they cancel out and are negligible. That way, you only have to worry about the y direction. Don't forget the force that gravity has the object. It appears to me that the object is falling, so there would be an additional force from going down from weight of the object. Weight is gravity (can be rounded to 10) x mass. Substitute 4N+weight in for Fnet and 1kg in for m.
(4N + 10 x 1kg)=(1kg)a
14/1=14, so the acceleration is 14 m/s²
