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Lesechka [4]
2 years ago
12

Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu

me that Phobos's orbit is circular. Calculate the mass of Mars. Express your answer in units of kg. (Hint: Use the circular orbit velocity formula ; remember to use units of meters, kilograms, and seconds.) Please round the answer to four significant digits. m
Physics
1 answer:
Elenna [48]2 years ago
4 0

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

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John has a boat that will travel at the rate of 15 kph in still water. he can go upstream for 35 km in the same time it takes to
MA_775_DIABLO [31]
Let F = the downstream speed of the water. 

<span>Then the boat's upstream speed is: 15 - F </span>
<span>The boat's downstream speed is: 15 + F </span>


<span>Assume both the journeys mentioned take T hours, then using "speed x time = distance" we get: </span>

<span>Downstream journey: (15 + F)T = 140 </span>
<span>Upstream journey: (15 - F)T = 35 </span>


<span>Add the two formulae together: </span>

<span>(15 + F)T + (15 - F)T = 140 + 35 </span>

<span>15T + FT + 15T - FT = 175 </span>

<span>30T = 175 </span>

<span>T = 35/6 </span>


<span>Use one of the equations to find F: </span>

<span>(15 + F)T = 140 </span>

<span>15 + F = 140/T </span>
<span>F = 140/T - 15 </span>
<span>F = 140/(35/6) - 15 </span>
<span>F = 24 - 15 </span>
<span>F = 9 </span>

<span>i.e. the downstream speed of the water is 9 kph </span>

<span>Therefore, the boat's speed downstream is 15 + F = 15 + 9 = 24 kph.
the answer is:                       *24kph*</span>
5 0
3 years ago
a worker uses a board that is 7 m long to pry up a bolder A small rock is used for the fulcrum and is placed 2.5 m from the resi
e-lub [12.9K]

Answer:

en español por

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4 0
2 years ago
A black squirrel is sitting on the ground. A red squirrel is sitting on a tree branch high above the black squirrel. Both squirr
labwork [276]

The black squirrel has zero kinetic energy (if it's not moving) and lower gravitational potential energy than the red squirrel or zero gravitational potential energy if the ground is assumed to be zero gravitational potential line.

4 0
2 years ago
Read 2 more answers
A cyclist accelerates from a velocity of 10 miles/hour east until reaching a velocity of 20 miles/hour east in 5 seconds. What w
Sliva [168]

Answer:

a = 0.894\ m/s^2

Explanation:

<u>Motion with Constant Acceleration</u>

A body moves with constant acceleration when the speed changes uniformly in time. The equation used to find the final speed vf is

v_f=v_o+at

Where vo is the initial speed, a is the acceleration, and t is the time.

The cyclist has an initial speed of vo=10 miles/hour and ends up at vf=20 miles/hour in t=5 seconds.

Both speeds are given in miles/hour and we must convert it to m/s:

1 mile/hour = 0.44704 m/s

10 mile/hour = 4.47 m/s

20 mile/hour = 8.94 m/s

The acceleration is calculated by solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

\displaystyle a=\frac{8.94-4.47}{5}

a = 0.894\ m/s^2

3 0
2 years ago
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther
jeka94

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = \frac{m}{\frac{I_o}{r}  +mr} \ r v_o

let's calculate

       v = \frac{0.020}{\frac{1.4}{0.6  } + 0.020 \ 0.6  } \ 0.6 \ 4

       v = \frac{0.020}{2.345} \ 2.4

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

6 0
2 years ago
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