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arlik [135]
1 year ago
9

A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l

owest floor
Physics
1 answer:
Lelechka [254]1 year ago
7 0

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

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irina1246 [14]

Answer:

House and Senate.

Explanation:

declares war, regulates interstate and foreign commerce and controls taxing and spending policies.

4 0
2 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × \sqrt{\frac{m}{k} }   ................2

put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0
3 years ago
A mature thunderstorm will contain both updrafts and downdrafts. True or False
jasenka [17]

A mature thunderstorm will contain both updraft and downdrafts. The given statement is true.

When the cumulus cloud becomes very large, the water in it becomes large and heavy. Raindrops start to fall through the cloud when the rising air can no longer hold them up. Meanwhile, cool dry air starts to enter the cloud. Because cool air is heavier than warm air, it starts to descend in the cloud (known as a downdraft). The downdraft pulls the heavy water downward, making rain.

This cloud has become a cumulonimbus cloud because it has an updraft, a downdraft, and rain. Thunder and lightning start to occur, as well as heavy rain. The cumulonimbus is now a thunderstorm cell.

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3 years ago
Read 2 more answers
A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his
pantera1 [17]

\Large {{ \sf {Question :}}}

<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

\Large {{ \sf {Given :}}}

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

\Large {{ \sf {Formulae  :}}}

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3>a \:  =  \frac{v - u}{t}</h3><h3>\Large {{ \sf {Step-by-step explanation :}}}</h3>

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\Large {{ \sf {Answer :}}}

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Varvara68 [4.7K]

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similarly kinetic friction on it

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so here the weight of the suspended block is less than the limiting friction on it

So here we will say that friction will counter balance the weight of the suspended block and it will not move at all

So acceleration of the box will be zero

5 0
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