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sergij07 [2.7K]
3 years ago
7

A mechanic has a wrench where the hand grip is 0.40 m from the axis of the bolt. To apply a torque of 170 N m with this wrench w

hile pushing at right angles to the wrench, the mechanic must exert a force of a. 0.0024 N b. 68. N c. 1.7 x 10 N d. 4.3 x 10 N e. None of these
Physics
1 answer:
Grace [21]3 years ago
4 0

Answer:

4.3 * 10 N

Explanation:

To calculate torque, we multiply the distance from the pivot by the perpendicular (the part of the force that acts at right angles to the displacement vector) component of the force to the displacement vector from the pivot.

torque  = distance from pivot *  perpendicular force

170 Nm= 0.4 m * F

F = 425 N = 4.3 * 10 N rounded off to two significant figures

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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
A skateboarder with a mass of 45 kilograms is riding on a skateboard with a mass of 2.5 kilograms. What should be the velocity o
klasskru [66]
So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.

Since we know momentum and mass we can now solve.

Velocity = 264/(45+2.5)
              = 5.56 m/s
5 0
3 years ago
Read 2 more answers
An auditorium measures 43.0m × 18.0m × 15.0m. The density of air is 1.20 kg/m3. What is the volume of the room in cubic feet? (D
frez [133]
The question here would be what is the volume of the room. The density of air that is given has no use. We simply multiply the dimensions given of the room to determine  the volume.

<span>43.0m × 18.0m × 15.0m = 11610m^3 ( 3.28 ft / 1 m)^3 = 4.09 x 10^5 ft^3</span>
5 0
3 years ago
An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t
elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

\mu = \frac{f_0}{f_e}

Here,

\mu = Magnification

f_e = Focal length eyepieces

f_0 = Focal length of the Objective

Rearranging to find the focal length of the objective

f_0 = \mu f_e

Replacing with our values

f_0 = 7.5* 3.7cm

f_0 = 27.75cm

Therefore the focal length of th eobjective lenses is 27.75cm

5 0
3 years ago
Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)
rewona [7]

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

3 0
3 years ago
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