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Alika [10]
3 years ago
13

If you run a 500 W computer for 4 hours, your electric meter will read ____; this is a measure of _______ "used"

Physics
2 answers:
emmasim [6.3K]3 years ago
7 0

Answer:

If you run a 500 W computer for 4 hours, your electric meter will read 2kWh; this is a measure of energy "used". The correct answer is D.

Explanation: Energy is defined as the capacity of doing work and is occurs in different forms( electrical, kinetic, potential energy). Electrical energy consumption can be calculated or measure through the multiplication of the power( W or kW) used by the number of hours during which the power is consumed. Therefore the unit of energy is kWh. From the question,

500W was used for 4 H

Therefore energy consumed

500× 4= 2000Wh. To convert to kWh

Divide by 1000,

2000÷ 1000= 2.

There energy consumed in kWh is,

2kWh energy.

Katen [24]3 years ago
3 0

Answer:

D) 2 kWh, energy

Explanation:

Energy = power×time

For electricity,

E = Pt.................... Equation 1

Where E = Electric Energy, P = power, t = time.

Given: P = 500 W, t = 4 hours.

Substitute into equation 1

E = 500(4)

E = 2000 Wh

E = (2000/1000) kWh

E = 2 kWh.

Hence the right option is D) 2 kWh, energy

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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
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Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

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