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mash [69]
3 years ago
10

A student pulls a box across a horizontal floor at a constant speed of 4.0 meters per second by exerting a constant horizontal f

orce of 45 newtons. approximately how much work does the student do against friction in moving the box 5.5 meters across the floor?
Physics
2 answers:
ruslelena [56]3 years ago
8 0
Work = Force multiplied by the distance(or displacement)
Rainbow [258]3 years ago
5 0

Answer:

247.5N

Explanation:

A student pulls a box across a horizontal floor at a constant speed of 4.0 meters per second by exerting a constant horizontal force of 45 newtons. approximately how much work does the student do against friction in moving the box 5.5 meters across the floor?

DEFINITION OF TERMS

work done is the product of force and distance. if the student pulls the box without covering a distance, e asnt done any form of work.

force is tat which tends to change a body's state of rest or uniform motion  in a straight line.

w=f*d

45*5.5

247.5Nm

the frictional force between the object and the ground will be in the opposite direction to the force applied

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A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
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Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

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