All categories

2 years ago

The Pickering nuclear power plant has a power rating of 3100 MW.

Physics

1 answer:

Roman55 [17]2 years ago

5 0

Answer:

$2.68\cdot 10^8 MJ$

Explanation:

The power is related to the energy by

$P=\frac{E}{t}$

where

P is the power

E is the energy

t is the time elapsed

The power of this nuclear power planet is

$P=3100 MW = 3.1\cdot 10^9 W$

The time we are considering is 1 day, which is

$t = 1 d = 86400 s$

So we can re-arrange the previous equation to find the energy produced by the power plant in one day:

$E=Pt =(3.1\cdot 10^9 W)(86400 s)=2.68\cdot 10^{14} J=2.68\cdot 10^8 MJ$

You might be interested in

Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = $5.976\times 10^{24} kg$

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm = 1.496\times 10^{11} m[/tex]

1 giga meter = $10^{9} meter$

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

$F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}$

Force between Sun and particle:

$F'=G\frac{mm'}{(r-x)^2}$

Force on particle is equal:

F = F'

$G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}$

$\frac{x}{r-x}=\sqrt{333,000}$ = ±577.06

Case 1:

$\frac{x}{r-x}=577.06$

x = $1.49\times 10^{11} m=149.34 Gm$

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

$\frac{x}{r-x}=-577.06$

x = $1.49\times 10^{11} m=149.86 Gm$

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0

2 years ago

Sebuah benda dijatuhkan bebas dari ketinggian 200 m jika grafitasi setempat 10 m/s maka hitunglah kecepatan dan ketinggian benda

Pie

Please post in English so i or someone else can help you.

7 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly

Viefleur [7K]

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

7 0

2 years ago

Read 2 more answers

Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s

PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0

2 years ago