The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body.
P = W = mass x acceleration due to gravity
P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N
Solving for the static friction force (F),
F = P x c
F = (2.94 N) x 0.6 = 1.794 N
Therefore, the maximum force of static friction is 1.794 N.
If you mean S is the distance then it is true
Velocity = Distance / time
Answer:
Approximate escape speed = 45.3 km/s
Explanation:
Escape speed
Here we have
Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²
R = 1 AU = 1.496 × 10¹¹ m
M = 2.3 × 10³⁰ kg
Substituting
Approximate escape speed = 45.3 km/s
Answer:
I believe that the answer is d.
Explanation:
Because there is nothing to make the aircraft accelerate or decelerate, it is going to stay in constant motion with no acceleration.
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '. D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .--
They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
