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ryzh [129]
3 years ago
9

if you double the mass of an object while leaving the net force unchanged what is the result of the acceleration

Physics
1 answer:
butalik [34]3 years ago
3 0
Force = mass*acceleration
acceleration = Force/mass

If you double the mass then acceleration will be halved in order for the equation to still equal.

New acceleration = Force/(2)mass
New acceleration = (1/2)* Force/mass------->if you compare this to original acceleration equation above it is 1/2
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We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient
zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

5 0
3 years ago
If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
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Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

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3 years ago
Analyze how some insects are able to move around on the surface of a lake or pond
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A stone is projected from the ground with a velocity 14m/s one second later it clears a wall 2 m high. The angle of projection i
Eva8 [605]

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3 years ago
Read 2 more answers
Simple physics question, check the document. Should take about 3-5 minutes.
Ahat [919]

Answer:

The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N

Explanation:

Forces on block 4.3 kg are:

63N to the right and R21 (contact force from the 6.3 kg block) to the left

Net force on 4.3 kg block is: 63 N - R21

Forces on the 6.3 kg block are:

R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.

So net force on the 6.3 kg block is: R12 - 11 N

According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").

Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:

a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2

solve for R by cross multiplication

6.3 (63 - R) = 4.3 (R - 11)

396.9 - 6.3 R = 4.3 R - 47.3

369.9 + 47.3 = 10.6 R

444.2 = 10.6 R

R = 444.2 / 10.6

R = 41.90 N

5 0
3 years ago
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