Can someone help me with this question please picture is above
1 answer:
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The final velocity of the skater is 2.34 m/s forward
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.
Before the ball is thrown, the total momentum is:
where
M = 61 kg is the mass of the skater
m = 4.0 kg is the mass of the ball
u = 2.5 m/s (forward) is the combined velocity of the skater and the ball
After, the ball is thrown at twice the velocity, so the final total momentum is
where
V is the final velocity of the skater
v = 2(2.5) = 5.0 is the final velocity of the ball
Since the total momentum must be conserved, we can write
So, the skater is moving at 2.34 m/s (forward) after the shot.
Learn more about momentum:
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This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( -
) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
