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Tom [10]
3 years ago
11

How does the force block A exerts on block B compare to the force block B exerts on block A?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

too much force

Explanation:

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The diagram below shows a 5.00-kilogram block
bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

  • <em>Mass of the block, m = 5 kg</em>

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

  • <em>g is acceleration due to gravity = 10 m/s²</em>

W = 5 x 10

W = 50 N <em>(</em><em>downwards</em><em>)</em>

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0
2 years ago
Read 2 more answers
A person travelled in the -x direction at a constant speed of 2.5 m/s. The person's final position is -10.0 m, and the initial p
Nostrana [21]

Answer:

(c) time required to travel = 8 sec

Explanation:

We have given the final position = -10 m on x axis

And the initial position =10 m

So total distance = 10-(-10)=20 m

The speed is given as 2.5 m/sec

We have tof ind the time required by the person to travel

Time is given by t=\frac{distance}{speed}=\frac{20}{2.5}=8 sec

So the option (c) is correct option

4 0
3 years ago
PLEASE HELP ME!<br><br> How do you think the swarms affect local populations of humans?
Volgvan

Answer:

by temperature

Explanation:

cuz

5 0
3 years ago
Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 11
Nataliya [291]

Answer:

0.182 m/s

Explanation:

m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s

let the velocity of loaded freight car is v

Use the conservation of momentum

m1 x u1 + m2 x 0 = (m1 + m2) x v

30,000 x 0.85 = (30,000 + 110,000) x v

v = 0.182 m/s

5 0
3 years ago
A bullet with a mass of 0.3 kg is fired out of a gun with a mass of 4 kg at 600 m/s. What is the recoil velocity on the gun?
slavikrds [6]

Answer:

According to the Conservation of Momentum,

Momentum of the gun = momentum of the bullet

M(gun)×V(gun)=m(bullet)×v(bullet)

4kg × V = 0.3kg × 600m/s²

V = (0.3 × 600)/4 = 45 m/s

The recoil velocity on the gun is <em><u>45 m/s</u></em>

<h3><u>45 m/s</u> is the right answer.</h3>
4 0
3 years ago
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