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777dan777 [17]
3 years ago
10

Since force is dp/dt, the force due to radiation pressure reflected off of a solar sail can be calculated as 2 times the radiati

ve momentum striking the sail per second. In the vicinity of Earth's orbit around the Sun, the energy intensity of sunlight is about 1300 W/m2. What is the approximate magnitude of the pressure on the sail? (For comparison, atmospheric pressure is about 105 N/m2.)
Physics
1 answer:
QveST [7]3 years ago
4 0

Answer:

8.67×10⁻⁶ N/m²

Explanation:

p = Momentum of a photon

E = Energy of a photon

c = Speed of light

I = Intensity of light

Force = dp/dt

p=\frac{E}{c}

\\\Rightarrow F=\frac{\frac{dE}{dt}}{c}

As given in question

F=2\frac{\frac{dE}{dt}}{c}

Now F/A = Pressure

P=\frac{2I}{c}\\\Rightarrow P=\frac{2\times 1300}{3\times 10^{8}}\\\Rightarrow P=8.67\times 10^{-6}\ N/m^2

∴ Magnitude of the pressure on the sail is 8.67×10⁻⁶ N/m²

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Answer:

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Explanation:

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2 years ago
Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0
ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

F_c=m\frac{v^2}{r}  (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

K=\frac{1}{2}mv^2

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}

you replace this value of v in the equation (1). Also, you replace the values of r and m:

F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N

hence, the tension in the string must be T =  Fc = 764.41 N

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3 years ago
A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict
Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

F=157\times 0.19\\\\F=29.83\ N

(B) The force is simply given by :

F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

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3 years ago
The magnetic flux through a loop:
dexar [7]

Answer:

The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.

Explanation:

Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.

It is expressed mathematically as

Φ = B A cos(θ) where

Φ is the magnetic flux

B is the magnetic field strength

A is the area

θ is the angle that the magnetic field make with the plane of the loop

If B is acting perpendicular to the plane of the loop, this means that θ = 90°

Magnetic flux Φ = BA cos90°

Since cos90° = 0

Φ = BA ×0

Φ = 0

This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.

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