What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal
2 answers:
Details are missing in the question. Complete text of the problem:
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"
Solution
(a) The pitcher accelerates the baseball from rest to a final velocity of , so
, in a time interval of
. The acceleration of the ball in the horizontal direction (x-axis) is therefore
And the distance covered by the ball during this time interval, before it is released, is:
(b) For this part we need to consider also the weight of the ball, which is
From this, we find its mass:
Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have
We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:
(upward)
So, the magnitude of the force is
To find the direction, we should find the angle of F with respect to the horizontal. This is given by
From which we find
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Answer:
a) perfectly inelastic, b) collision is inelastic, c) elastic
Explanation:
In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved
Initial instant. Before the skater touches the frisbee
p₀ = M v₁ + m v₂
where M and m are the masses of the skater and frisbee, respectively
for the final moment they give us several possibilities, in all case the moment is conserved
p₀ =
case a)
Final instant. grabs the frisbee and holds it
p_{f} = (M + m) v '
p₀ = p_{f}
We can see that this shock is perfectly inelastic, it holds the fressbee
case b)
final instant.
This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.
case c)
final instant. Grab the fressbee and resend it
this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.
Answer:
1:4
Explanation:
The formula for calculating kinetic energy is:
If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!