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tia_tia [17]
3 years ago
5

Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days

and the mass of the planet it orbits is 9.38E24 kg.You may assume the orbit to be circular.
Physics
1 answer:
Vinil7 [7]3 years ago
8 0

Answer:

The distance is r = 55430496 \  m  

Explanation:

From the question we are told that

   The period of the moon T =  1.2 days = 1.2 * 24 * 3600 = 103680 \  s

    The mass of the planet is  m_p =  9.38*10^{24} kg

Generally the period of the moon is mathematically represented as

       T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

Here G is the gravitational constant with value

        G  =  6.67 *10^{-11} \  N \cdot m^2/kg^2

=>   T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

=>   103680   =  2 *  3.142  * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }

=>    272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}

=>    r = \sqrt[3]{ 1.7031241*10^{23}}j

=>   r = 55430496 \  m        

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A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?
Ber [7]
U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
                   =(30-10)/6
                   =(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
                              =60+ 0.5*(10/3)*36
                              =120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
       =(900-100)/6.6666666.......
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7 0
3 years ago
how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?
ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0
3 years ago
which type of organism converts wastes and dead material into nutrients that can be used by plants (1) carnivore (2) herbivore (
Pavlova-9 [17]
Producer hope this helped
7 0
3 years ago
Can someone please help
Maru [420]

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5 0
3 years ago
) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick
sveticcg [70]

Answer:

  about 19.6° and 73.2°

Explanation:

The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...

  y = -4.9(x/s·sec(α))² +x·tan(α)

where s is the launch speed in meters per second.

We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):

  -13.6111·tan(α)² +50·tan(α) -16.0511 = 0

This has solutions ...

  tan(α) = 0.355408 or 3.31806

The corresponding angles are ...

  α = 19.5656° or 73.2282°

The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

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I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

5 0
2 years ago
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