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3 years ago

Please help! i'm horrible at this

Physics

1 answer:

frez [133]3 years ago

4 0

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

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PLEASE HELPPPP !!!! Scientific investigations must be well ____ to be sure the data collected will help answer the scientists qu

konstantin123 [22]

Answer-

Analyzed or conducted

Hope this helped

3 0

2 years ago

San Jose to San Francisco is approximately 7200 meters away. If it takes you 3600 seconds to drive by car what is the velocity?

svetlana [45]

V=s/t=>7200/3600=2m/s

5 0

2 years ago

You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of

Alecsey [184]

Answer:

18 newtons

Explanation:

Divide weight by speed

3 0

2 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

2 years ago

A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W

dalvyx [7]

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

$F_g = mg$

now the acceleration of ball is given as

$a = \frac{F_g}{m}$

$a = \frac{mg}{m} = g$

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

5 0

2 years ago