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garik1379 [7]
3 years ago
8

Question 11 (2 points)

Physics
1 answer:
____ [38]3 years ago
7 0

Answer:

B

Explanation:

Power is defined as the work done per unit time

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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
Lelechka [254]

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

7 0
2 years ago
The image shows a roller coaster.
mariarad [96]
I believe the answer is c but I’m not 100% sure
8 0
3 years ago
Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l
Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0
3 years ago
O operate this generator, a human applies a force by turning the crank. This action generates an electric current, which lights
Shtirlitz [24]
Answer A I’m very positive about it
3 0
3 years ago
Read 2 more answers
An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts
Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0
3 years ago
Read 2 more answers
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