Answer:
When the elevator is accelerating downward, the person feels lighter due to the downward normal force being less than the person's weight.
Explanation:
A person riding in an elevator subjected to a series of unbalanced forces depending on the direction the elevator is travelling.
Two forces are acting on the person; the force of gravity and the upward normal force from the elevator.
When the elevator is going upwards with acceleration a, the person feels heavier than his normal weight, due to the upward normal force being greater than the person's weight. N = mg + ma
When the elevator is moving downwards with acceleration a, the person feels lighter due to the downward normal force being less than the person's weight. N = mg - ma
However, when the elevator is moving up or down at constant velocity ie. acceleration a = 0, the person experience a normal force equal to weight. N = mg
When the elevator is moving downwards with acceleration a = g, the person experiences weightlessness. N = (mg - mg) = 0
Answer:
1.23×10⁸ m
Explanation:
Acceleration due to gravity is:
a = GM / r²
where G is the universal gravitational constant,
M is the mass of the planet,
and r is the distance from the center of the planet to the object.
When the object is on the surface of the Earth, a = g and r = R.
g = GM / R²
When the object is at height i above the surface, a = 1/410 g and r = i + R.
1/410 g = GM / (i + R)²
Divide the first equation by the second:
g / (1/410 g) = (GM / R²) / (GM / (i + R)²)
410 = (i + R)² / R²
410 R² = (i + R)²
410 R² = i² + 2iR + R²
0 = i² + 2iR − 409R²
Solve with quadratic formula:
i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)
i = [ -2R ± √(1640R²) ] / 2
i = (-2R ± 2R√410) / 2
i = -R ± R√410
i = (-1 ± √410) R
Since i > 0:
i = (-1 + √410) R
R = 6.37×10⁶ m:
i ≈ 1.23×10⁸ m
The number of protons
number of neutrons=the mass number- number of protons
14-6=8
Answer:
1. Torque → F. Study of forces
2. C.O.G → D. Point of action of weight.
3. Plumb line → A. Line of C.O.G
The impulse given to the ball is equal to the change in its momentum:
J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s
This is also equal to the product of the average force and the time interval ∆t :
J = F(ave) ∆t
so that if F(ave) = 200 N, then
∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s
