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Pepsi [2]
3 years ago
9

A 7.40-kg object initially has 347 J of gravitational potential energy. Then an elevator lifts the object a distance of 20.6 m a

bove its previous position. How much work did the elevator perform on the object?
Physics
1 answer:
storchak [24]3 years ago
6 0

Answer:

W = 1493.9 J = 1.49 KJ

Explanation:

The work done by the elevator on the object will be equal to the gain in is potential energy:

W = ΔP.E

W = mgΔh

where,

W = Work = ?

m = mass of object = 7.4 kg

g = 9.8 m/s²

Δh = gain in height = 20.6 m

Therefore,

W = (7.4 kg)(9.8 m/s²)(20.6 m)

<u>W = 1493.9 J = 1.49 KJ</u>

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The electricity received at an electrical substation has a potential difference of 20,000 V. What should the ratio of the turns
iren2701 [21]

The turn ratio of the transformer is 40.

<h3 />

<h3>What is a transformer?</h3>

A transformer is an electrical device that is used to step up or step down voltage.

The formula for calculating the turns ratio of a transformer is given below.

<h3> Formula</h3>
  • a = V/v.......... Equation 1

Where:

  • V = Primary voltage
  • v = Secondary voltage
  • a = Turns ratio of the transformer

From the question,

Given:

  • V = 20000 V
  • v = 500 v

Substitute the values above into equation 1

  • a = 20000/500
  • a = 40.

Hence, The turn ratio of the transformer is 40.

Learn more about transformers here: brainly.com/question/6450076

7 0
3 years ago
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-Dominant- [34]
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3 years ago
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If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu
zubka84 [21]

Answer:

minimum speed is 15.35 m/s

frictional coefficient  is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and   minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² =  236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 =  0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245  = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient  is 0.26

4 0
3 years ago
Read 2 more answers
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

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WINSTONCH [101]

Answer:

true is the answer of the question

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