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Kazeer [188]
3 years ago
8

A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s .

If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is:
Physics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

50 N

Explanation:

given,

mass of ball = 1 Kg

initial velocity = 3 m/s

final velocity = 2 m/s

time = 0.10 s

impulse = change in momentum

I = m ( v - u)            

I = 1( 2 -(-3))                

I = 5 kg.m/s          

Force is equal to impulse per unit time

average force = \dfrac{I}{t}

                       = \dfrac{5}{0.1}

                       = 50 N

Average force on the floor will be equal to 50 N

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What is the weight of a 1-kilogram brick resting on a table?
MakcuM [25]

Answer:

The weight if the block is 10Newtons

Explanation:

The weight of any object is quantity of matter the object contains and it is always acting downwards on such body. This shows that the object is under the influence of gravity.

The weight of an object is calculated as mass of the object × its acceleration due to gravity

W = mg

Give the mass of the brick to be 1kg

g is the acceleration due to gravity = 10m/s²

Weight of the object = 1 × 10

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5 0
3 years ago
Is 2/2 1 or 0? please help lol
Komok [63]

Answer:

1.

Explanation:

Hello!

In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:

2 ÷ 2 = 1.

Best regards!

4 0
3 years ago
Jupiter has the strongest gravity of any planet in our solar system. Which quantity will be a maximum there?
Mamont248 [21]

weight will be maximum there

3 0
2 years ago
If you pull horizontally on a desk with a force of 150 N and the desk doesn't move, the friction force must be 150 N. Now if you
s344n2d4d5 [400]

Answer:

The friction force is 250 N

Explanation:

The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):

\sum F=ma

In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:

\sum F = 0 (1)

We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:

- the pull, forward, F = 250 N

- the friction force, backward, F_f

Given (1), we have

F-F_f = 0

So the force of friction must be equal to the pull:

F=F_f = 250 N

8 0
3 years ago
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The total amount of energy stays the same, but throughout the ride, the kinetic energy and the potential energy change, still adding up to the same number. At the top of the ride it has potential energy, and as it goes down the potential energy decreases and the kinetic energy increases. When it’s at the bottom of the first drop it has maxed out its kinetic energy, and minimized its potential energy. Friction slows down the car, and pushes on the cart with a force that is equal and opposite to the force being exerted in the track. The reason the track keeps going is because though it exerts and equal and opposite force the momentum of the objects is different, allowing the car to continue moving, however friction will slow it down until eventually it comes to a stop.
8 0
3 years ago
Read 2 more answers
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