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Ivan
2 years ago
13

Determine the work done by the constant force. The locomotive of a freight train pulls its cars with a constant force of 15 tons

a distance of one-quarter mile.
Physics
1 answer:
Digiron [165]2 years ago
5 0

Answer:

5.92×10⁷ J

Explanation:

We'll begin by converting 15 tons to Newton. This can be obtained as follow:

1 ton = 9806.65 N

Therefore,

15 ton = 15 ton × 9806.65 N / 1 ton

15 ton = 147099.75 N

Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:

100 mi = 160934 m

0.25 mi = 0.25 mi × 160934 / 100 mi

0.25 mi = 402.335 m

Finally, we shall determine the Workdone. This can be obtained as follow:

Force (F) = 147099.75 N

Distance (d) = 402.335 m

Workdone (Wd) =?

Wd = F × d

Wd = 147099.75 × 402.335

Wd = 5.92×10⁷ J

Thus, the Workdone is 5.92×10⁷ J

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8 0
3 years ago
If there is a 25 [lbs] of force acting on a 5 [lbs] of mass, what is the acceleration of that mass?
mihalych1998 [28]

Answer:

f = 25 lbs

m = 5 lbs

a =?

f = ma

25 = 5 a( divide both sides by 5)

a = 5(lbs)

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2 years ago
A leopard of mass 65kg climbs 7m up a tall tree. Calculate how much gravitational potential energy it gains. Assume g=10N/kg.
vesna_86 [32]

Answer: 4550 Joules

The formula for potential energy in gravitational field is as follows:

E_{pot} = mgh=65kg \cdot 10 \frac{N}{kg} \cdot 7m = 4550 J

8 0
3 years ago
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
1 year ago
A drop of oil of volume 10m it spread out on water to make a circular firm of radius 10m calculate the tickness of the firm
Effectus [21]

Answer:

h = 3.1 cm

Explanation:

Given that,

The volume of a oil drop, V = 10 m

Radius, r = 10 m

We need to find the thickness of the film. The film is in the form of a cylinder whose volume is as follows :

V=\pi r^2 h\\\\h=\dfrac{V}{\pi r^2}\\\\h=\dfrac{10}{\pi \times 10^2}\\\\h=0.031\ m\\\\h=3.1\ cm

So, the thickness of the film is equal to 3.1 cm.

8 0
2 years ago
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