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3 years ago

(6) Module 4 Study Guide

Physics

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velikii [3]3 years ago

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We have the meats Arby’s we beat them kids

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A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

3 years ago

Explain how forces of attraction and repulsion exist within an atom

Stels [109]

Answer:

Oppositely charged particles attract each other, while like particles repel one another. Electrons are kept in the orbit around the nucleus by the electromagnetic force, because the nucleus in the center of the atom is positively charged and attracts the negatively charged electrons.

Explanation:

7 0

3 years ago

Read 2 more answers

Why might you want to use a single fixed pulley?

hjlf

A single fixed pulley can be used to raise or lower lightweight objects.

Option b

<u>Explanation:</u>

A pulley is a simple machine tool which is used to make lifting or lowering tasks easy. A single fixed pulley is a system involving only one pulley fixed on a constant rigid support with a rope wrapped around the wheel. Such a system can be used only to change the direction of applied force in raising or lowering small, lightweight objects which need minimal work force.

A single fixed pulley system helps only in redirecting the applied force direction by using a rope and wheel assembly. The work done in such a case remains the same and hence it is not preferred to use it in lifting heavy objects. Neither is the required force reduced in case of a single fixed pulley system. A movable pulley helps in achieving (A) and (C).

4 0

3 years ago

A car moves in a straight line such that for a

jeyben [28]

Answer:

s = 90 m

a = 56 m/s²

Explanation:

I will ASSUME that your equation is silly as it reduces to V = 11t which is constant, and that you mean V = 9t² + 2t

Position is the integral of differential velocities

s = $\int\limits^3_0 {9t^2 + 2t} \, dt$

s = 3t³ + t² | from 0 to 3

s = 3(3)³ + 3² - (0) = 90 m

acceleration is the derivative of velocity

a = v' = 18t + 2

a(3) = 18(3) + 2 = 56 m/s²

3 0

3 years ago

What can quantum physics tell us about the nature of our reality?

Alenkinab [10]

As do pixels create such a defined picture of virtual reality it would be similar to say that such small particles (with understanding and exploring them) would show us how things interact and that basic interaction might show us the result of the creating of something from which we can discover by observing the final result of that partical or even as if looking at a cake- to soon know what it takes to make a cake maybe by it's flavor or how it's not yet known "ingredients" interact with the known ingredients and that interaction shows that ingredients to the cake and the ingredients might as well be the future of that cake

7 0

3 years ago