M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.
The angular speed is
ω = 280 rpm
= (280 rev/min)*(2π rad/rev)*(1/60 min/s)
= 29.3215 rad/s
The moment of inertia of the bug is
I = mr²
= (2.2 x 10⁻³ kg)*(0.03 m)²
= 1.98 x 10⁻⁶ kg-m²
Calculate the angular momentum of the bug.
J = Iω
= (1.98 x 10⁻⁶ kg-m²)*(29.3215 rad/s)
= 5.806 x 10⁻⁵ (kg-m²)/s
Answer: 5.806 x 10⁻⁵ (kg-m²)/s
Potential energy = mgh
So, energy gained
= mgh
= 70kg × (9.8m/s²) × 1000m
= 686000 kgm²/s²
= 686000 J
(a)
The formula is:
∑ F = Weight + T = mass * acceleration
as the elevator and lamp are moving downward, I choose downward forces to be
positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2
(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides
(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63
mass = 63 / 11.5
Mass = 5.48 kg
(b)
Since the elevator and lamp are going upward, I choose upward forces to be
positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7
-53.70 + T = 5.48 * 1.7
T = 53.70 + 9.316 = approx 63 N
The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated
upward at the same rate of 1.7 m/s^2
Answer:
6.9%
Explanation:
We are given that
Distance between Earth and Moon=d=
m
Radius of Earth=r=
a.Initially gravitational pull
.....(1)
After changing
Gravitational pull
.....(2)
Equation (2) divided by equation (1)
Increases in gravitational pull=
%
