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Ksenya-84 [330]
3 years ago
7

A block slides on a frictionless surface. The block is given a 25 N force to the right and it accelerates at a rate of 3.5 m/s^2

to the right. This block is then placed on a frictionless surface on Jupiter and given a force of 15 N. What is the acceleration on the block? (1 point)
1.7 m/s^2
8.9 m/s^2
4.3 m/s^2
2.1 m/s^2
Physics
2 answers:
otez555 [7]3 years ago
6 0

<u>Answer:</u> The acceleration of the block is 2.10m/s^2

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=m\times a       .......(1)

where,

F = force exerted on the block = 25 N

m = mass of the block = ?

a = acceleration of the block = 3.5m/s^2

Putting values in equation 1, we get:

25kg.m/s^2=m\times 3.5m/s^2\\\\m=\frac{25}{3.5}=7.14kg

As, the mass of the block remains the same everywhere in the universe.

Now, calculating the acceleration on the block on Jupiter by using equation 1:

Force on the block = 15 N

Mass of the block = 7.14 kg

Putting values in equation 1, we get:

15kg.m/s^2=7.14kg\times a\\\\a=\frac{15}{7.14}=2.10m/s^2

Hence, the acceleration of the block is 2.10m/s^2

sasho [114]3 years ago
5 0

Answer:

2.1 m/s^2

Explanation:

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I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s.
AfilCa [17]

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

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Substituting and re-arranging the equation, we can find the final velocity of the ball:

v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s


2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

m_1 u_1 + m_2 u_2 = (m_1 +m_2) v

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s


3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

I=\Delta p (1)

The impulse can be rewritten as product between force and time of collision:

I=F \Delta t

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

\Delta p = m\Delta v = m(v_f -v_i)

So, eq.(1) becomes

F \Delta t = m(v_f -v_i)

where:

F = ? is the unknown force

\Delta t = 0.002 s is the duration of the impact

m = 0.16 kg is the mass of the ball

v_f = 44 m/s is the final velocity of the ball

v_i = -38 m/s is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

F=\frac{m (v_f -v_i)}{\Delta t}=\frac{(0.16 kg)(44 m/s-(-38 m/s))}{0.002 s}=6560 N

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vladimir2022 [97]

Answer:

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