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Leokris [45]
4 years ago
9

A 60.0-kg person is in an elevator with a mass of 2000 kg. The elevator moves vertically up through a distance of 24.0 m with a

constant speed of 4.00 m/s. How much work is done on this elevator system? Use g=10.0m/s^2.
Physics
1 answer:
kati45 [8]4 years ago
6 0

The mass of the (elevator + person) is (2,000 kg + 60 kg) = 2,060 kg .

The weight is (mass x gravity) = (2,060 x 10) = 20,600 newtons

Work = (force x distance) = (20,600 newtons x 24m) = <em>494,400 joules</em>

The speed, acceleration, and how much time it takes don't make any
difference, unless you want the average power during the lift.

         Power = (work)/(time) =
           494,000 J / (24/4 m) =
           494,000 J / 6 sec = <em><u>82,400 watts</u></em>      wow !
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3 years ago
Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is 1.5732*10^{-11}V

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

\epsilon_{max} = NBA\omega

Where,

N= Number of turns of the coil

B = Magnetic field

\omega = Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

\omega = 2\pi f

\omega = 2\pi*61.5

\omega =123\pi rad/s

Replacing at the equation our values given we have that

\epsilon_{max} = NBA\omega

\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega

\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)

\epsilon_{max} = 1.5732*10^{-11}V

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is 1.5732*10^{-11}V

6 0
4 years ago
A single loop of wire with an area of 0.0920 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
jekas [21]

Answer:

Induced emf in the loop is 0.02208 volt.

Induced current in the loop is 0.0368 A.

Explanation:

Given that,

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The initial value of uniform magnetic field, B = 3.8 T

The magnetic field is decreasing at a constant rate, \dfrac{dB}{dt}=0.24\ T/s

(a) The induced emf in the loop is given by the rate of change of magnetic flux.

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\times \dfrac{dB}{dt}\\\\\epsilon=0.092\times 0.24\\\\\epsilon=0.02208\ V

(b) Resistance of the loop is 0.6 ohms. Let I is the current induced in the loop. Using Ohm's law :

\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{0.02208}{0.6}\\\\I=0.0368\ A

Hence, this is the required solution.

5 0
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8 0
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