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shtirl [24]
2 years ago
9

A(n) 636 kg elevator starts from rest. It moves upward for 4.5 s with a constant acceleration until it reaches its cruising spee

d of 2.05 m/s. The acceleration of gravity is 9.8 m/s 2 . Find the average power delivered by the elevator motor during this period. Answer in units of kW.
Physics
1 answer:
zmey [24]2 years ago
3 0

Answer:

The average power delivered by the elevator motor during this period is 6.686 kW.

Explanation:

Given;

mass of the elevator, m = 636 kg

initial speed of the elevator, u = 0

time of motion, t = 4.5 s

final speed of the elevator, v = 2.05 m/s

The upward force of the elevator is calculated as;

F = m(a + g)

where;

m is mass of the elevator

a is the constant acceleration of the elevator

g is acceleration due to gravity = 9.8 m/s²

a = \frac{v-u}{t} \\\\a = \frac{2.05 -0}{4.5} \\\\a = 0.456 \ m/s^2

F = (636)(0.456 + 9.8)

F = (636)(10.256)

F = 6522.816 N

The average power delivered by the elevator is calculated as;

P_{avg} = \frac{1}{2} (Fv)\\\\P_{avg} = \frac{1}{2} (6522.816 \ \times \ 2.05)\\\\P_{avg} = 6685.89 \ W\\\\P_{avg} =  6.68589 \ kW\\\\P_{avg} =  6.686 \ k W

Therefore, the average power delivered by the elevator motor during this period is 6.686 kW.

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A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block
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Answer: mg/Cosθ

Explanation:

Taking horizontal acceleration of wedge as 'a'

FCosΘ = FsinΘ

F = mass(m) × acceleration(a) = ma

For horizontal resolution g = 0

Therefore,

Horizontal = Vertical

maCosΘ = mgSinΘ

aCosΘ = gSinΘ

a = gSinΘ/CosΘ

Recall from trigonometry :

SinΘ/Cosθ = tanΘ

Therefore,

a = gtanΘ

Normal force acing on the wedge:

mgCosΘ + maSinΘ - - - - (y)

Substitute a = gtanΘ into (y)

mgCosΘ + mgtanΘsinΘ

tanΘ = sinΘ/cosΘ

mgCosΘ + mgsinΘ/cosΘsinΘ

mgCosΘ + mgsin^2Θ/cosΘ

Factorizing

mg(Cosθ + sin^2Θ/cosΘ)

Taking the L. C. M

mg[(Cos^2θ + sin^2Θ) /Cosθ]

Recall: Cos^2θ + sin^2Θ = 1

mg[ 1 /Cosθ]

mg/Cosθ

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3 years ago
Which of the following is the newton used to measure? answer
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3 years ago
An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?
Cloud [144]

The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

v²= 2×10.8 m/s²×400m

v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

To know more about speed

brainly.com/question/13489483

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