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3 years ago

14

An escalator is used to move 25 passengers every minute from the first floor of a department store to the second. The second flo

or is located 5.30 meters above the first floor. The average passenger's mass is 60 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time.
Please help me quickly!

2 answers:

kompoz [17]3 years ago

8 0

Answer:

55960 J

Explanation:

Anit [1.1K]3 years ago

5 0

Answer:

1298.5watt

Explanation:

Power is defined as the rate at which work is done

Expressed mathematically as;

P = F × d/ t; F = force or weight of object ;d is height or distance sustain and t is time covered

Hence for 1 person the power consumption is ;

Note weight =mass× acceleration of free fall due to gravity,g

g is known as 9.8m/S2

Time taken is 1min which is 60 sec

60 ×9.8 × 5.3/60 = 51.94 watt

Hence for 25 persons we have

25 × 51.94 =1298.5watt

1.3kilo watt( kilo is 1000)

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A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp

AlekseyPX

Answer:

(A) 2.4 N-m

(B) $0.035kgm^2$

(C) 315.426 rad/sec

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(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

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Force F = 20 N

(a) Torque is equal to product of force and distance

So torque $\tau =Fr$ , here F is force and r is distance

So $\tau =20\times 0.12=2.4Nm$

(B) Moment of inertia is equal to $I=\frac{1}{2}mr^2$

So $I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2$

Torque is equal to $\tau =I\alpha$

So angular acceleration $\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2$

(C) As the disk starts from rest

So initial angular speed $\omega _{0}=0rad/sec$

Time t = 4.6 sec

From first equation of motion we know that $\omega =\omega _0+\alpha t$

So $\omega =0+68.571\times 4.6=315.426rad/sec$

(D) Kinetic energy is equal to $KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J$

(E) From second equation of motion

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad$

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3 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

Vaselesa [24]

Answer:

$1.57 * 10^{3} Q$

Explanation:

The volume charge density is defined by ρ = $\frac{Q}{V}$ (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are $\frac{Coulombs}{m^{3} }$ , so we have to express the radius in meters:

inner radius = $4 cm * \frac{1 m}{100 cm} = 0.04m$

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Now, we know that the volume of the sphere is calculated by the formula:

$V = \frac{4}{3}\pi r^{3}$ , and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = $\frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3})$ , where $r_{o}$ is the outer radius and $r_{i}$ is the inner radius.

Replacing the volume formula in the Equation A:

ρ = $\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}$

ρ = $\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }$

Replacing the values of the outer and inner radius whe have:

ρ = $\frac{3Q}{4\pi (1.52 * 10^{-4})}$

ρ = $1.57 * 10^{3} Q$

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