A person approaches a flat mirror at a speed of 0.25 m / s. How fast is he approaching his image?
1 answer:
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a) The acceleration of gravity is
b) The critical velocity is 6668 m/s (24,006 km/h)
c) The period of the orbit is 8452 s
d) The satellite completes 10.2 orbits per day
e) The escape velocity of the satellite is 9430 m/s
f) The escape velocity of the rocket is 11,191 m/s
Explanation:
a)
The acceleration of gravity for an object near a planet is given by
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
h is the height above the surface
In this problem,
(mass of the Earth)
(Earth's radius)
(altitude of the satellite)
Substituting,
b)
The critical velocity for a satellite orbiting around a planet is given by
where we have again:
(mass of the Earth)
(Earth's radius)
(altitude of the satellite)
Substituting,
Converting into km/h,
c)
The period of the orbit is given by the circumference of the orbit divided by the velocity:
where
v = 6668 m/s
Substituting,
d)
One day consists of:
While the period of the orbit is
T = 8452 s
So, the number of orbits completed by the satellite in one day is
e)
The escape velocity for an object in the gravitational field of a planet is given by
where here we have:
Substituting, we find
f)
We can apply again the formula to find the escape velocity for the rocket:
Where this time we have:
, because the rocket is located at the Earth's surface, so its altitude is zero.
And substituting,
Learn more about gravitational force:
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Given :
A box weighing 12,000 N is parked on a 36° slope.
To Find :
What will be the component of the weight parallel to the plane that balances friction.
Solution :
The component of that will be parallel to the plane to balance friction is :
Therefore, component of force to balance friction is F sin 36° .
Hence, this is the required solution.
The capacitance is defined as the maximum charge stored in a capacitor, Q, divided by the voltage applied, V:
The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula:
The capacitor is initially charged with the battery of 108 V, so the the initial charge on the capacitor can be found by re-arranging the previous formula: