Answer:
A. min
Explanation:
= mass of earth = 6.0 x 10²⁴ kg
= Radius of earth = 6400 km = 6.4 x 10⁶ m
= Altitude above earth = 400 km = 0.4 x 10⁶ m
Radius of orbit of the space station around earth is given as
km
m
= Time period of orbit of space station
Here we can use Kepler's third law which is given as
Inserting the above values
sec
The frictional force is 218.6 N
Explanation:
The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.
There are two forces acting along this direction:
- The component of the weight parallel to the incline, downward along the plane, of magnitude
where
m = 46 kg is the mass
is the acceleration of gravity
is the angle of the incline
- The (static) frictional force, acting upward, of magnitude
Since the block is in equilibrium, we can write
And substituting, we find the force of friction:
Learn more about frictional force along an inclined plane:
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