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Sedaia [141]
4 years ago
15

Estimate the average power output of the Sun, given that

Physics
1 answer:
Mrac [35]4 years ago
5 0

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

I=\frac{P}{A}

Where P is the power through an area A.  

Rearranging to find the Power we have that,

P = IA

The value of the given area for land can be approximated to that of a sphere, therefore the Area would be equivalent to

A = 4\pi r^2

Where r represents the distance from the sun to the earth.

Replacing with our values we have that

P = (1350W/m^2)(4\pi r^2)

P =1350*(4\pi (1.5*10^{11})^2)

P = 38151*10^{22} W

Therefore the average power output of the sun is P = 38151*10^{22} W

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What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a
andrezito [222]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

6 0
4 years ago
The basic unit of mass in the International System of Units, or SI, is the a. meter. c. liter. b. ounce. d. gram.
Elodia [21]
The basic unit of mass in the International System of Units. or SI, is D. gram. 
7 0
3 years ago
A parked car begins to roll down a hill, what can you conclude from that observation?
Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

3 0
3 years ago
Select the correct answer. You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour
BlackZzzverrR [31]

Answer:

Since velocity is a vector quantity there is no net displacement and the average velocity is zero

(A) is correct

7 0
2 years ago
6. A plane due to fly from Montreal to Edmonton required refueling. Because the fuel gauge on the aircraft was not working, a me
LiRa [457]

Answer:

The amount of liters of fuel should have been added is  20093 L

Explanation:

Given that;

a mechanic used a dipstick to determine that 7682 L of fuel were left on the plane

i.e the volume of fuel left in the plane = 7682 L

Required fuel to make  a trip = 22,300 kg of fuel

Also from the question; we are being told that in order for the pilot to determine the volume ; he asked for the density of the fuel and the mechanic said 1.77.

This volume of fuel was added and the plane subsequently ran out of fuel, but landed safely by gliding into Gimli Airport near Winnipeg. The error arose because the factor 1.77 was in units of pounds per liter (lbs/L).

Now; we can understand that the density of the fuel was 1.77 pound /litre.

SO , let convert 1.77 pound /litre to kg/Litre;

we all know that

1 pound = 0.4536 kg

1.77 pound/litre  = x kg

If we cross multiply ; we will have:

1.77 pound/litre  × 0.4536 kg = 1 pound × x kg

x kg = (1.77 pound/litre  × 0.4536 kg) /1 pound

x = 0.802872 kg/litre

\mathbf{Density = \dfrac{mass}{volume}}

where ;

mass =  22,300 kg of fuel

volume = unknown ???

density = 0.802872 kg/litre

making volume the subject of the formula from above; we have:

\mathbf{volume = \dfrac{mass}{Density}}

\mathbf{volume = 22300 \  kg \ of \ fuel *\dfrac{1 \ litre }{0.802872 \  kg \ of \ fuel}}

volume = 27775.28672 litre

volume \approx 27775 L

Let not forget that we are being told as well that the volume of fuel left in the plane = 7682 L

Now;

The amount of liters of fuel should have been added is: =  27775 L - 7682 L

The amount of liters of fuel should have been added is  20093 L

8 0
3 years ago
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