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4 years ago

1. The picture below shows Jamal pushing with a 100-Newton (N) force on a large box. Neither

Physics

2 answers:

posledela4 years ago

5 0

Answer:

<h2>100 newtons</h2>

Explanation:

r-ruslan [8.4K]4 years ago

4 0

Answer:

100 newtons

Explanation:

Given,

Jamal pushing a large box by a force, F = 100 N

Work done on the large box is, W = 0

It is because the applied force is less than the force of the friction between the two surfaces.

Yet, there will be a force that is exerted by the large box on Jamal.

According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

$F_{a} =-F_{r}$

Hence, If the action force is 100 N, then the reaction force should be in 100 N

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Two charges are separated by a distance 'd' and exert a mutual attractive force of 'f' on each other. if the distance is decreas

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Peanut butter an jelly crackers are the best

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3 years ago

The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

$r(t)=(8t+9)i+(2t^2-8)j+6tk$

Let v is the velocity of the particle. Velocity of an object is given by :

$v=\dfrac{dr(t)}{dt}$

$v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}$

$v=(8i+4tj+6k)\ m/s$

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

$a=\dfrac{dv(t)}{dt}$

$a=\dfrac{d[8i+4tj+6k]}{dt}$

$a=(4j)\ m/s^2$

At t = 0, $v=(8i+0+6k)\ m/s$

$v(t)=\sqrt{8^2+6^2} =10\ m/s$

Hence, this is the required solution.

7 0

3 years ago

HELP PLZ ASAP!!!!!

zubka84 [21]

Answer:

C. Oxygen combines with carbon dioxide

Explanation:

B i o l o g y

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3 years ago

If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?

andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

$x=4-12t+3t^2$

$\frac{dx}{dt}=-12+6t$

For t = 1:

$\frac{dx}{dt}=-6$

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3 years ago

A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no

Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

When brougth close to the charged sphere A, as charges can move freely in a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
If in such state, if the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of charge principle, no net charge can be built on sphere C.

3 0

3 years ago