Answer:
work = 1125 [J]
Explanation:
To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:
Power = work/time
Power = 12.5 [w]
work = joules [J]
time = 1.5 [min] = 90 [s]
work = 12.5*90
work = 1125 [J]
Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
Answer:
F₁ = 4,120.2 N
F₂ = 3,924N
Explanation:
1) Balance of angular momentum around the end where F₁ is applied.
F₂ × 0.5m - F₁ × 0 = mass × g × 1m
⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m
F₂ = 196.2 Nm / 0.5m = 3,924 N
2) Balance of forces
F₁ - F₂ = mg
F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N
Answer:
r = 9.92 mm
Explanation:
Given that,
Mass of oil drop, 
It acquires 2 surplus electrons, q = +2e 
Potential difference, V = 620 V
Thie potential difference is applied between the pair of horizontal metal plates the drop is in equilibrium.
We need to find the distance between the plates.
At equilibrium,
mg = qE
Since, E = V/r (r is distance between plates)
So, the distance between the plates is 9.92 mm.
