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2 years ago

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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]2 years ago

5 0

Ans;

6.25m²/Kd

Explanation: see attached file

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A 2-column table with 4 rows. The first column labeled substance has entries calcium chloride, calcium bromide, calcium carbonat

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Answer:

SAMPLE C

Explanation:

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A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has

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Answer:

Explanation:

Given

mas of block $m=4\ kg$

speed of block $v=2.5\ m/s$

spring constant $k=30\ N-m$

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$x=v\sqrt{\frac{m}{k}}$

$x=2.5\times \sqrt{\frac{4}{30}}$

$x=0.912\ m$

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Planet A has mass 3M and radius R, while Planet B has mass 4M and radius 2R. They are separated by center-to-center distance 8R.

Aleksandr-060686 [28]

Answer:

Explanation:

In Newton's law of universal gravitation

F = Gm₁m₂/r²

Where G is a gravitational constant = 6.674e-11m³/kgs²

m₁ and m₂ are the masses of the two bodies or objects in question, in kilogram (kg)

r is the distance in meters between them

From the question, the rock is placed halfway between the planets

So, it's distance from planet A is 8R/2 = 4R

And it's distance from planet B is also 8R/2 = 4R

Using F = Gm₁m₂/r²

To Planet A

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet A = 3M

So Fa = G mm₂/r² = Gm(3M) / (4R)²

To Planet B,

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet B = 4M

Fb = G mm₂/r² = Gm(4M) / (4R)²

Comparing both forces together, we realise that Planet B has the largest force,

so take we F = Fb – Fa

F = Fb – Fa = Gm(4M) / (4R)² – Gm(3M) / (4R)²

F = GmM/16R²)(4–3)

F = GmM/16R²

Note that Force = Mass * Acceleration

So, F = ma

So, ma = GmM/16R² ------- Divide through by m

a = GM/16R²

From the question

M = 7.3×10^23kg

R = 5.8×10^6 m

So, a = (6.674 * 10^-11 * 7.3×10^23)/16(5.8×10^6)²

a = (48.7202 * 10^12)/16(33.64 * 10^12)

a = (48.7202 * 10^12)/(538.4 * 10^12)

a = 48.7202/538.4

a = 0.090517612960760

a = 0.091m/s² ----------Approximated

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A ray of light is traveling from water into a diamond. Upon hitting the diamond, a light ray bends toward the normal. With this

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The correct answer is b

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En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

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Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

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