Answer:
Acceleration, 
Explanation:
Given that,
Height from a ball falls the ground, h = 17.3 m
It is in contact with the ground for 24.0 ms before stopping.
We need to find the average acceleration the ball during the time it is in contact with the ground.
Firstly, find the velocity when it reached the ground. So,

u = initial velocity=0 m/s
a = acceleration=g

It is in negative direction, u = -18.41 m/s
Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

So, the average acceleration of the ball during the time it is in contact is
.
Answer:
9.8 m/s2
Explanation:
In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.
Got it from the internet, hope it helps though ^^
Answer:
19 m/s
Explanation:
The complete question requires the final speed to be calculated.
Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.
For this question, first the unknown acceleration must be calculated and used to determine the final velocity
Step 1: Calculate the acceleration




Step 2: Calculate the velocity using the acceleration calculated above



In 6 secs, the dog covers-
S=vt
8.9*6 = 53.4 m.
In the same time, the cat covers, 53.4-3.8 = 49.6 m.
Thus, speed of the cat, v= s/t,
= 49.6/6 = 8.267 m/s
Answer:
Decreased by a factor of 4.5
Explanation:
"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

where
is the gravitational constant on Earth.
are the masses of the object and Earth itself. and R distance between, or the Earth radius.
So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:




Since
and 

So gravity would have been decreased by a factor of 4.5