Answer:
point of support on which a lever rotates.
Explanation:
The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.
The fulcrum provides the platform for a lever to torque.
- The force that opposes motion by the applied force is termed the frictional force.
- Friction is a force that opposes motion.
- The stored energy of an object is its potential energy.
- The potential energy is the energy due to the position of a body.
- The distance an object moves when doing work is termed its displacement.
Torque = r x F
|F| = mg = 60 * 10 N = 600 N ( assuming g ~ 10m/s^2)
distance of fulcrum = torque / Force = 90/600 m = .15 m.
Answer:
6.86 × 10²⁴ kg
Explanation:
The mass of the earth m = density of earth, ρ × volume of earth, V
m = ρV
The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m
Since m = ρV
m = ρ4πr³/3
So, substituting the values of the variables into the equation for the mass of the earth, m, we have
m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3
m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3
m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3
m = 6546105.64378π × 10¹⁸ kg/3
m = 20565197.400122 × 10¹⁸ kg/3
m = 6855065.8 × 10¹⁸ kg
m = 6.8550658 × 10²⁴ kg
m ≅ 6.86 × 10²⁴ kg
An equilibruium is not changed by a changed in pressure. the answer is false
Answer:
5.3 m/s
Explanation:
First, find the time it takes for him to fall 7m.
y = y₀ + v₀ t + ½ at²
0 = 7 + (0) t + ½ (-9.8) t²
0 = 7 − 4.9 t²
t ≈ 1.20 s
Now find the velocity he needs to travel 6.3m in that time.
x = x₀ + v₀ t + ½ at²
6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²
v₀ ≈ 5.27 m/s
Rounded to two significant figures, the man must run with a speed of 5.3 m/s.
