Answer:
Acceleration=4m/s²
Force applied =619.8N
Explanation:
Using equation of motion
V=u+at we have: u=o, v=50m/s
50= 0 + a×0.0121
a = 50/0.0121
a= 4m/s²
Neglecting resistance forces
F= ma, where a = v-u/t
F=m×(v-u)/t
F= 0.150 ×(50-0)/0.0121
F=7.5/0.0121
F= 619.8N
Hi there!
We know that:
U (Potential energy) = mgh
We are given the potential energy, so we can rearrange to solve for h (height):
U/mg = h
g = 9.81 m/s²
m = 30 g ⇒ 0.03 kg
0.062/(0.03 · 9.81) = 0.211 m
There are two forces acting on a parachute with a parachutist: the force of gravity and the air resistance. The force pulling the skydiver to the ground would be Wright force
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = 
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = 
(|r₂₁|)² = 148.25
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Answer:
4,524,660 N
Explanation:
Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.
m/9 = (1026 kg/m³) (50 m³)
m = 461,700 kg
mg = 4,524,660 N
