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jok3333 [9.3K]
3 years ago
11

If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 52.0 m/s in the opposite directi

on, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. Express your answer with the appropriate units.

Physics
1 answer:
stich3 [128]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

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A baseball is thrown directly upward from ground level with a velocity of +15 m/s. What are the two times when the ball is 10 m
erastovalidia [21]

Answer:

time is 0.5660 s

and time is - 3.62431  s

Explanation:

velocity u = 15 m/s

height s = 10 m

acceleration due to gravity g =  –9.8 m/s²

to find out

time

solution

we will apply here distance equation that is

s = ut - 1/2× gt²   ...........1

here put all these value and get time t

here s is height and g is -9.8

so

s = ut - 1/2× gt²

10 = 15t - 1/2× (-9.8)t²

10 = 15t + 4.9t²

solve it we get t

t = 0.56630 and -3.62431

so time is 0.5660 s

and time is - 3.62431  s

8 0
4 years ago
A runner starts at position A. He runs 40 m North, 10 m East and 40 m
Alecsey [184]
The answer is C.) 10 m East
8 0
3 years ago
Laura puts her file cabinet on a small cart with wheels which will occur. A Laura will increase the amount of friction. B Laura
kow [346]

Answer: b

Explanation:

6 0
3 years ago
Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer
sveticcg [70]

by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

\theta = 180 - 54.3 - 79.1 = 46.6 degree

now using above equation

93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6

4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

4 0
3 years ago
NEED HELP ASAP PLEASE
Zarrin [17]

Answer:

D. 3.0 m/s

Explanation:

because I did this in my class

8 0
3 years ago
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