Answer:
The coefficient of static friction is : 0.36397
Explanation:
When we have a box on a ramp of angle
, and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.
In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

and the force of static friction (f) is given as the static coefficient of friction (
) times the normal N:

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.
In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.
Remember that the Force of Gravity is given under the principle

Where,
G = Gravitational Universal constant
M = Mass of the planet
m = mass of the object
r = Distance from center of the planet
When the radius grows considerably the gravitational force begins to decrease.
Answer:
- The impulse exerted by the ball is 7.2 kg.m/s
- The average force exerted on the ball by the bat is 600 N.
Explanation:
Given;
mass of the baseball, m = 0.144 kg
velocity of the baseball, v₁ = 20 m/s
velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)
The impulse exerted by the ball is calculated as follows;
J = ΔP = mv₁ - mv₂
ΔP = m(v₁ - v₂)
ΔP = 0.144 [20 - (-30)]
ΔP = 0.144 ( 20 + 30 )
ΔP = 0.144 (50)
ΔP = 7.2 kg.m/s
The average force exerted on the ball by the bat is calculated as;

1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
= 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11 AU
= 3.0.74 / 100 = 0.0374 AU