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Amiraneli [1.4K]
4 years ago
15

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

verted to energy. How much mass is converted per hour if the efficiency of the plant is 30 percent? (Answer is in kilograms)
Physics
1 answer:
alukav5142 [94]4 years ago
6 0

Answer:

0.00016 kg

Explanation:  

Given:

Power = P = 1.2 × 10⁹ Watts

Power =  work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 =  4  × 10⁹ W

Energy =  4  × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c²  = (1.44 x 10¹³) / (3 × 10⁸ )²

                                   = 0.00016 kg

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Mary pushes a crate by applying force of 18 newtons. Unable to push it alone, she gets help from her friend, Anne. Together they
maw [93]
 <span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is: 

Ff = UsN 

where Us is the coefficient of static friction and N is the normal force. 

In order to get the crate moving you must first apply enough force to overcome the static friction: 

Fapplied = Ff 

Since Fapplied = 43 Newtons: 

Fapplied = Ff = 43 = UsN 

and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11 

43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
7 0
3 years ago
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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

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Answer:

D. Dissolve

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3 0
3 years ago
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