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Artist 52 [7]
3 years ago
14

If electrical energy costs 7¢/kW•h, calculate

Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

10 minutes =(1/6)th of an hour

cost of running it for an hour = 0.09*7=0.56 c

in 10 min the cost should be (1/6)*0.56c=0.08c

Explanation:

Your scanner power is 15W so, it will consume 15*60=90Wh of electricity in an hour. that is 0.09KWh.  

So, cost of running it for an hour = 0.09*7=0.56 c

Which means that in 10 min the cost should be (1/6)*0.56c=0.08c

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If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
Ilya [14]
Work of the force = 10 N

Time required for the work = 50 sec

Height = 7 m

We are given with the value of work and time in the question.

Substitute the values in the formula of power and then you'll get the power required.

We know that,

w = Work

p = Power

t = Time

By the formula,

Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,

p = 70/50

p = 1.4 watts

Therefore, the power required is 1.4 watts.

Hope it helps!
3 0
3 years ago
In the following diagram, the voltage is 1.5 volts and the resistance is 7.5 ohms. Use Ohm's law to determine the current in the
Basile [38]

I believe it is 0.2 amps  Hope this helps sorry if wrong

5 0
3 years ago
Read 2 more answers
‏A 50 - N x m torque acts on a wheel with a moment of inertia 150 kg x m² . If the wheel starts from rest , how long will it tak
denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, \triangle \theta = 2\pi

Torque, \tau = 50 Nm

Moment of Inertia, I = 150 kg m^2

If the wheel starts from rest, w_{0} = 0 rad/s

The angular displacement of the wheel can be given by the formula:

\triangle \theta = \omega_0 t + 0.5 \alpha t^2................(1)

Where \alpha is the angular acceleration

\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2

To get t, put all necessary parameters into equation (1)

2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s

3 0
3 years ago
Can some one help me with this question its hard :( down below ill give brainliest
eduard

Answer:  B

Explanation:

8 0
3 years ago
Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea
steposvetlana [31]

Answer

given,

V = 2 L

the left is an ideal gas at  P = 100 k Pa and T = 500 K

mass is constant

 m_1 = m_2

\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}

Pressure is same because it's not changing due to process

\dfrac{V}{500} = \dfrac{2 V}{T_2}

T_2 = 1000\ K

\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}

\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})

m = \dfrac{P_1V_1}{RT_1}

m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}

m = 1.39 x 10⁻³ Kg

\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)

\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K

5 0
3 years ago
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