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2 years ago

Action and reaction forces will cancel because they are pushing on the same object toward each other. True or False

Physics

2 answers:

sweet-ann [11.9K]2 years ago

5 0

Answer:

False

Explanation:

Newton's third law states that for every action there is an equal and opposite reaction. The action-reaction pair act on different bodies. Thus, they do not cancel each other.

For example, in order to walk, we push the floor backwards. This is action force which the foot applies to the floor. The floor pushes the foot forward. This is the reaction force. The amount of force is equal but the direction is opposite. Also, the bodies on which action force and reaction force are acting are different.

Hence, the given statement is false.

vampirchik [111]2 years ago

4 0

Sometimes. If the forces are equal, they will cancel out and the net force will be zero. But if the forces are not equal, the object will move away from the stronger force because there is still a net force.

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Which is capeble of housing astronaughts while they conduct reasearch

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You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

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If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

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2 years ago

Some people wish that we lived in a recollapsing universe that would eventually stop expanding and start contracting. For this t

Monica [59]

Answer:

(B) Dark energy does not exist and there is much more matter than current evidence suggests.

Explanation:

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3 years ago

Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is

NISA [10]

Answer: 0.006in/s

Explanation:

Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s

Also let the rate at which the radius of the balloon is increasing be dr/dt

Given r = 4.7in and Π = 3.14

Applying the chain rule method

dV/dt = dV/dr × dr/dt

If the volume of the sphere is 4/3Πr³

V = 4/3Πr³

dV/dr = 4Πr²

If r = 4.7in

dV/dr = 4Π(4.7)²

dV/dr = 277.45in²

Therefore;

1.68 = 277.45 × dr/dt

dr/dt = 1.68/277.45

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2 years ago

shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r

Ainat [17]

Answer:

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0$

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$\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0$

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

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2 years ago

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