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3 years ago

10

Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a

rea density for the mirror at which the radiation pressure exactly cancels out the gravitational attraction from the sun?

1 answer:

Sati [7]3 years ago

5 0

Ans;

6.25m²/Kd

Explanation: see attached file

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A 1210 kg rollercoaster car is

ratelena [41]

Answer: 4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

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3 years ago

The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the

lara [203]

To hit the target the crew drop the crate before the plane is directly over the target. It is because <span>because the cargo has forward velocity and therefore before it reaches the ground it travels some distance. The answer is A. Hope it helps. </span>

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3 years ago

Read 2 more answers

A light pointer is stuck to the rubber sheet so that it pivots about a point P near the

deff fn [24]

I’m not going to church tomorrow or Friday I don’t want to go go back up

8 0

2 years ago

Why does ice reflect more energy compared to water?

Stells [14]

Answer:

while ice is made by water again it melts and becomes water. water is colourless and odourless and has no taste but ice is only cold and hard. water is used for drinking and other things. but is for freshness and it never flows

Explanation:

so ice reflect more energy compared to water

3 0

3 years ago

Suppose a car of mass m is moving at a constant speed v of

SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

$\arcsin(v^2/(gR))$

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

$w_r = mg\sin(\theta)$

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

$F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g$

Also, the car is making a circular motion:

$a = \frac{v^2}{R}$

Combining the equations:

$\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}$

Finally the angle is

$\arcsin(v^2/(gR))$

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3 years ago