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denpristay [2]
2 years ago
9

Two boxes sit side by side on a smooth horizontal surface. The lighter box 5.2 kg, the heavier box has a mass of 7.4kg (a) find

the contact force between these boxes when a horizontal force of 5.0 N is applied to the light box. (Hint: you need to calculate acceleration first)
Physics
1 answer:
Paladinen [302]2 years ago
8 0

OK.  So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.

-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.

-- You're pushing it with 5.0N of force.

-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)

-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.

The force on it to make it accelerate is (mass) x (acceleration) =

                                                              (7.4 kg) x (5/12.6 m/s²) =  <em>2.936 N.</em>

The only force on the big box comes from the small box, pushing it from behind. 
So that same  <em>2.936N</em>  must be the contact force between the boxes.

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Jules is conducting an experiment involving friction. He is measuring the temperature of various objects and surfaces after quic
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Answer:

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A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
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2 years ago
Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.
dimulka [17.4K]

Answer:

The distance of separation is d = 0.092 \ m

Explanation:

The mass of the each ball is  m= 10 g  =  0.01 \ kg

 The negative charge on each ball is q_1 =q_2=q =  1 \mu C  =  1 *10^{-6} \ C

Now we are told that the lower ball is  restrained from moving this implies that the net force acting on it is  zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

          F =  \frac{kq_1 * q_2}{d}

=>       m* g  =  \frac{kq_1 * q_2}{d}

here k the the coulomb's  constant with a value  k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

So  

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5 0
2 years ago
A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the
egoroff_w [7]

The acceleration of the jet is 28.1 m/s^2

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2

Learn more about acceleration:

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