Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
Answer:
i think it's weakest
EDIT: It's net. I answered weakest but it was wrong and the correct answer was net. oops
Explanation:
if a strong force is acting on something it will push it away, meaning the object would go towards the weaker force
First, we would need to know the decaying isotope.
Next, we use the decay formula
A = Ao e^(-kt)
After determining the remaining amount after two hours, the decay reaction can be used to determine the number of gamma rays released. If the given is in terms of mole, then the total energy is
E = 140n KeV where n is the number of moles of gamma rays released
The answer is refracts parallel to the axis of the lens
