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3 years ago

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Two electric motors drive two elevators of equal mass in a three-story building 10 meters tall. Each elevator has a mass of 1,00

0 kg.
The second elevator takes 10 seconds to do this work.

Calculate the power output of the second motor.

9.8 x 102 W
9.8 x 103 W
9.8 x 104 W
9.8 x 105 W

2 answers:

Afina-wow [57]3 years ago

8 0

The power output of second motor is 9.8 $\times$ 10^3 W.

<u>Explanation:</u>

Power is defined as the rate of doing electric work. I short it can be viewed as

Power = work done / time

Work done is the product of mass, gravity, and height. (i.e. mgh)

Given that m = 1000 kg, h = 10 m, g = 9.8 m / s^2, t = 10 s

Power = work done / time

= (m $\times$ g $\times$ h) / time

= (1000 $\times$ 9.8 $\times$ 10) / 10

power = 9800 W = 9.8 $\times$ 10^3 W.

fenix001 [56]3 years ago

8 0

Answer:

9.8 \times 10^3 W

Explanation:

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Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the

Leni [432]

Answer:

Part(a): The relative capacitance is $\dfrac{C_{A}}{C_{B}} = 0.33$

Part(b): The relative energy stored is $\dfrac{U_{A}}{U_{B}} = 0.33$

Part(c): The relative charge stored is $\dfrac{Q_{A}}{Q_{B}} = 0.33$

Explanation:

We know the capacitance ( $C$ ) of a capacitor having charge ( $Q$ ) and subjected to a potential difference of ( $V$ ) is given by

$C = \dfrac{Q}{V}$

Also, the energy ( $U$ ) stored by a capacitor can be written as

$U = \dfrac{1}{2}C~V^{2}$

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be $\textbf{r_{i}^{B}}$ $\bf{r_{i}^{B}}$ , the outer radius be $\bf{r_{o}^{B}}$ , the inner radius of Capacitor A be $\bf{r_{i}^{A}}$ and the outer radius be $\bf{r_{o}^{B}}$ .

Given in the problem,

$&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}$

Now, the capacitance ( $C$ ) of a cylindrical capacitor is given by,

$\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}$

where $\epsilon_{o}$ is the permittivity of the free space, $L$ is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

$C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}$

and the capacitance of capacitor B,

$C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}$

giving the relative capacitance of each capacitor to be

$\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33$

Part(b):

Energy stored by capacitor A,

$U_{A} = \dfrac{1}{2}~C_{A}~V^{2}$

Energy stored by capacitor B,

$U_{B} = \dfrac{1}{2}~C_{B}~V^{2}$

giving the relative energy stored by each capacitor to be

$\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

Part(c):

The charge stored by capacitor A,

$Q_{A} = C_{A}~V$

The charge stored by capacitor B,

$Q_{B} = C_{B}~V$

giving the relative charge stored by each capacitor to be

$\dfrac{Q_{A}}{Q_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33$

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If the ice absorbed 350,000 joules in 5 minutes, then it absorbed energy
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julsineya [31]

Explanation:

Yes, evidently, the process is consistent with the scientific process because in scientific process falsification and modification are two very important traits. So this new concept have modified the existing theories.

Through the modification a theory is adjusted without undermining other discovering made through the theory's prediction.

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IRISSAK [1]

Answer:

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u = 10 cosθ

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x= 5 m

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x = u .t

5 = 5 .t

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Vo= 10 sin 60°

Vo = 8.66 m/s

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$0=V_o^2-2gh$

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Mkey [24]

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