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.
Answer:
1362000 kgm/s
Explanation:
So the total mass combination of the plane and the people inside it is
M = 35000 + 160*65 = 45400 kg
After 15 seconds at an acceleration of 2 m/s2, the plane speed would be
V = 2*15 = 30 m/s
So the magnitude of the plane 15s after brakes are released is
MV = 45400 * 30 = 1362000 kgm/s
Answer:
D
Explanation:
Because we have two reactants and product
Answer:
I think its structural plasticity.
(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in
(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s
(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:
(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s
(c) The plane accelerates toward the center of the path with magnitude
<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²
(d) By Newton's second law, the tension in the line is
<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N
