Question is missing. Found on google:
<em>"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. </em>
<em>Part B
</em>
<em>What is the net force on the ball during the hit? </em>
<em>Express your answer to two significant figures and include the appropriate units."</em>
Solution:
A) 
The acceleration of the ball is given by
where
v = 12 m/s is the final velocity
u = 0 is the initial velocity (the ball is stationary)
t = 2.0 ms = 0.002 s is the time of contact
Substituting,
B) 
The force on the ball can be found by using Newton's second law:
where
m = 140 g = 0.14 kg is the mass of the ball
is the acceleration
Substituting,
A hypothesis is a proposed explanation made on the basis of limited information while a theory is a series of ideas intended to explain something.
First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:
2as = v² - u²
Because the final velocity v is 0 in such cases
s = -u²/2a; because both u and a are downwards, the negative sign cancels
s = 14.5² / 2*9.81
s = 10.72 meters
Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m
We will use the formula
s = ut + 0.5at²
to find the time taken with the initial velocity u = 0.
55.72 = 0.5 * 9.81 * t²
t = 3.37 seconds
Answer:
Explanation:
average speed more than 25.0m/s.
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to
