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3 years ago

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Many spacecraft have visited Mars over the years. Mars is smaller than the earth and has correspondingly weaker surface gravity.

On Mars, the free-fall acceleration is only 3.8 m

1 answer:

sertanlavr [38]3 years ago

7 0

Incomplete question.The complete one is here

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?

Answer:

$T=5900s=99min$

Explanation:

Given

$r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\$

To find

orbital period of a spacecraft T

Solution

An the initial calculating is computing the angular velocity of satellite :

$w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s$

Computing T

$T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min$

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Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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4 years ago

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You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the

Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

$t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2}$

T(final) = 500* 10 + 100*70/600 = 20°C

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3 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i

stiks02 [169]

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s² (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

<u>P.E = 0.068 J = 68 mJ</u>

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3 years ago

A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.6

andriy [413]

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

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Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

$2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}$

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

$m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6$ ,

hence, the mass of the rod is 126.6 g.

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3 years ago

This same experiment is performed on the international space station. What is the primary issue with performing this experiment

lions [1.4K]

Answer:

Difference in experimental data.

Explanation:

There is difference of experimental value between the experiment that is performed on the earth and on the international space station because presence of gravity. The result of the experiment on the earth is different due to the presence of gravity that contributes in the result of the experiment as compared to international space station where no gravity is present so there is high difference of the numerical value of the result of both experiments of earth and international space station.

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