Answer:
because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!
Answer:
<h2>1.17 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula

f is the force
m is the mass
From the question we have

We have the final answer as
<h3>1.17 m/s²</h3>
Hope this helps you
Answer:
ANSWER BELOW I
I
V
Remember that w=mg where w is weight in Newtons, m is mass in kilograms, and g is gravity in
m/s2
. For example, for Earth, 445 N = 45.4 × 9.8
m/s2
:Notice that the x-axis values will be gravity in
m/s2
, which is already given in the table, and the y-axis values will be the weight in Newtons. Remember to round your weights to a whole number, and to enter the points starting with the lowest gravity (moon, then Mars, then Venus, then Earth).
The efficiency of an ideal Carnot heat engine can be written as:

where

is the temperature of the cold region

is the temperature of the hot region
For the engine in our problem, we have

and

, so the efficiency is
Answer:

Explanation:
Given data:
first case
Distance of electric field from center of sphere is r_1 <R
Electric field at r_1< R

second case
Distance of electric field from centre of sphere is r_1 < 2R
Electric field at r_1< 2R

so, we have
