Ok so the first thing you need to do is im not gonna help you
Answer:
0.2
Explanation:
The given parameters are;
The acceleration of the train, a = 0.2·g
The mass of the person standing on the train = m
Let μ represent the coefficient of static friction, we have;
The force acting on the person, F = m × a = m × 0.2·g
The force of friction acting between the feet and the floor,
= m·g·μ
For the person not to slide we have;
The force acting on the person = The force of friction acting between the feet and the floor
F = 
∴ m × 0.2·g = m·g·μ
From which we get;
0.2 = μ
The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
is the velocity of Bill with respect to Amy (which is stationary)
is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

Therefore, Carlos velocity in Bill's reference frame will be

So, the magnitude is 20 m/s and the direction is westward (negative sign).
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa