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Ipatiy [6.2K]
3 years ago
15

Number 29 plz help physics

Physics
1 answer:
svetoff [14.1K]3 years ago
8 0

Answer:

a. E = 900000 J = 900 KJ

b. ΔT = 8.18 °C

c. Cost = $ 7.2  

Explanation:

a.

The energy can be given by:

E = Pt

where,

E = Energy = ?

P = Power = 500 W

t = time = (0.5 h)(3600 s/1 h) = 1800 s

Therefore,

E = (500\ W)(1800\ s)

<u>E = 900000 J = 900 KJ</u>

b.

The change in temperature of room air is given by the following formula:

0.5E = mC\Delta T\\   (since 50% of energy is used to heat air)

where,

m = mass of air = 50 kg

C = specific heat of air = 1.1 KJ/kg.°C

ΔT = Change in temperature of air = ?

Therefore,

(0.5)(900\ KJ) = (50\ kg)(1.1\ KJ/Kg.^oC)\Delta T

<u>ΔT = 8.18 °C</u>

c.

First, we will calculate the total energy consumed:

E = (0.5\ KW)(6\ h/d)(30\ d)\\E = 90\ KWh

Now, for the cost:

Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(90\ KWh)

<u>Cost = $ 7.2</u>

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Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

r_1 = Distance between center of mass of person and pole = 0.45 m

r_2 = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s

The new angular speed is 1.70344 rad/s

7 0
3 years ago
A weight lifter lifts a 345 N set of weights from ground level to a position over his head, a vertical distance of 1.89 m. How m
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Answer:652.05 J

Explanation:

Given

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vertical distance move h=1.89 m

Work done in lifting the weight is equal change in Potential Energy of weight

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\Delta PE=345 \times 1.89=652.05 J

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3 0
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In a thundercloud there may be an electric charge of 41 C near the top of the cloud and −41 C near the bottom of the cloud. If t
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F = 236063.6N

Explanation:

Please see attachment below.

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If someone were monitoring your vital signs (like your heart rate, oxygen content in your blood, blood pressure, etc.) every hou
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Answer:

Accuracy

Explanation:

I think accuracy is more important. When it comes to vital organs in the body, the exactness of getting the measurement is paramount. Accuracy deals with getting very close, almost exact you may say, to a known standard. Precision on the other hand, deals with how easy a measurement can be retaken, reproduced or remade, irrespective of how far or close they are from the accepted norm.

From this, we can agree that precision neglects the most important factor, closeness or say, exactness. Precision isn't bothered by it. And while that can be excused in a few instances, it certainly can not be permitted when it comes to life, or organs of the body

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3 years ago
The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
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Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

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The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

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Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

3 0
3 years ago
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