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2 years ago

four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?

2 answers:

7 0

If you and I were playing " tug-a-war " and we both pull on opposite sides with the same amount of force no one would move, the rope would be still, because the same force is on both sides of it.

Kay [80]2 years ago

3 0

There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.

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A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

5 0

1 year ago

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and

natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\\\\\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s$

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.

4 0

1 year ago

A box with a mass of 100.0 kg slides down a ramp with a 50 degree angle. What is the weight of the box? N What is the value of t

nadya68 [22]

1) weight of the box: 980 N

The weight of the box is given by:

$W=mg$

where m=100.0 kg is the mass of the box, and $g=9.8 m/s^2$ is the acceleration due to gravity. Substituting in the formula, we find

$W=(100.0 kg)(9.8 m/s^2)=980 N$

2) Normal force: 630 N

The magnitude of the normal force is equal to the component of the weight which is perpendicular to the ramp, which is given by

$N=W cos \theta$

where W is the weight of the box, calculated in the previous step, and $\theta=50^{\circ}$ is the angle of the ramp. Substituting, we find

$N=(980 N)(cos 50^{\circ})=630 N$

3) Acceleration: $7.5 m/s^2$

The acceleration of the box along the ramp is equal to the component of the acceleration of gravity parallel to the ramp, which is given by

$a_p = g sin \theta$

Substituting, we find

$W_p = (9.8 m/s^2)(sin 50^{\circ})=7.5 m/s^2$

5 0

1 year ago

Which of the following forces exists between objects even in the absence of direct physical contact

den301095 [7]

Answer: TRUST ME I GOT IT WRONG the answer is B

Explanation:

3 0

2 years ago

Read 2 more answers

Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl

Trava [24]

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac = q ( Va - Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]

= Wac - Wab

= Wac + 3Wac

= 4Wac

4 0

1 year ago