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3 years ago

Hannah has information about an object in circular orbit around Earth.

Physics

2 answers:

Svetllana [295]3 years ago

4 0

The distance of the orbiting object to Earth will give Hannah the radius which she needs.

Elena-2011 [213]3 years ago

3 0

The correct answer is:

the distance of the orbiting object to Earth.

In fact, we know that the gravitational force that keeps the object in circular motion around the Earth is equal to the centripetal force, so we can write:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

If we re-arrange the equation, we find an expression for the tangential speed of the object:

$v=\sqrt{\frac{GM}{r}}$

and we see that it depends on 3 quantities: G, M (the mass of the Earth) and r (the distance of the object from the Earth).

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A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is

OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

$U = \frac{GMm}{r}$

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

$U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}$

$U_2= - 5.1180*10^9J$

Point B )

We now use the average radius distance from the earth.

$U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}$

$U_2= -5.8925*10^9J$

Then,

$\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)$

$\Delta U = 774.5*10^6$

By the law of conservation of energy we know that,

$\Delta U = \frac{1}{2}mv^2$

clearing v,

$v= \sqrt{2 \Delta U/m}$

$v= \sqrt{2*774.5*10^6 /94.2}$

$v= 4055.08m/s$

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0

3 years ago

If we've traveled with medium and displaces the particles perpendicular to the direction in which the wave is traveling. This ty

AlexFokin [52]

This type of wave is a transverse wave, or an S wave (secondary wave)

4 0

3 years ago

Which of the following statements is accurate? A) Compressions and rarefactions occur throughout a transverse wave. B) The wavel

Nikitich [7]

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave

5 0

3 years ago

Assume that the operating cost of a certain truck (excluding driver's wages) is 12+x/6 cents per mile when the truck travels at

podryga [215]

Answer:

$x = 60 mph$

Explanation:

Given that the operating cost is

$c = 12 + \frac{x}{6}$ cents per mile

total miles covered is given as

$d = 400 miles$

so total cost of drive is given as

$C = (12 + \frac{x}{6})(4)$ $

time taken by the truck to move the distance is given as

$t = \frac{400}{x}$

So total earnings of the driver is given as

$E = \frac{400}{x} \times 6$ $

now total profit of the driver is given as

$P = \frac{2400}{x} - (48 + \frac{2x}{3})$ $

to maximize the profit we have

$\frac{dP}{dx} = 0$

$-\frac{2400}{x^2} + \frac{2}{3} = 0$

so we have

$x = 60 mph$

8 0

3 years ago

You launch a model rocket that has a mass of 1.5 kg. At a height of 300 m, it is traveling at 125 m/s. What is it's kinetic ener

Komok [63]

Answer:

KE = 11,719 J

Explanation:

KE = ½ mv²

KE = ½ (1.5 kg) (125 m/s)²

KE = 11,719 J

5 0

3 years ago