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3 years ago

11

If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?

1 answer:

Tema [17]3 years ago

5 0

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)

Divide each side by (mass):

   Acceleration = (force) / (mass)

   = (100 N) / (50 kg)

   = 2 m/s²

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What happens to most of the light waves that strike a clear pane of glass

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A person lowers a bucket into a well by turning the hand crank, as the drawing illustrates. The crank handle moves with a consta

dimulka [17.4K]

Answer:

0.453 m/s

Explanation:

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3 years ago

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

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$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

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Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

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Answer:

Answer for the question is given in the attachment.

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