Answer:
60 ÷ 5= 12 is the speed, hope it helps
With mass 2500 kg, the elevator has weight
(2500 kg) <em>g</em> = (2500 kg) (9.80 m/s²) = 24,500 N
The net force on the elevator is 30,000 N - 24,500 N = 5500 N (because its weight pulls it downward, while the tension pulls upward).
The acceleration <em>a</em> of the elevator is obtained from Newton's second law,
(2500 kg) <em>a</em> = 5500 N
<em>a</em> = (5500 N) / (2500 kg)
<em>a</em> = 2.2 m/s²
Answer:
a) Andrea's initial momentum, 200 kg m/s
b) Andrea's final momentum, 0
c) Impulse, = - 200 Ns
d) The force that the seat belt exerts on Andrea, - 400 N
Explanation:
Given data,
The initial velocity of the car is, u = 40 m/s
The mass of Andrea, m = 50 kg
The time period of deceleration, a = 0.5 s
The final velocity of the car, v = 0
a) Andrea's initial momentum,
p = mu
= 50 x 40
= 200 kg m/s
b) Andrea's final momentum
P = mv
= 50 x 0
= 0 kg m/s
c) Impulse
I = mv - mu
= 0 - 200
= - 200 Ns
The negative sign indicated that the momentum is decreased.
d) The force that the seat belt exerts on Andrea
F = (mv - mu)t
= (0 - 200) / 0.5
= - 400 Ns
Hence,the force that the seat belt exerts on Andrea is, - 400 N
Answer:
(a) 16.5528 V
(b) 827.64 W
Explanation:
(a)
The formula for maximum emf produced in a coil is given as
E₀ = BANω.................. Equation 1
Where E₀ = maximum emf produced in the coil, B = magnetic Field, A = Area of the rectangular coil, N = number of turns, ω = angular velocity.
Given: B = 1.9 T, N = 44 turns, ω = 110 rad/s, A = 2×9 = 18 cm² = 18/10000 = 0.0018 m²
Substitute into equation 1
E₀ = 1.9(44)(110)(0.0018)
E₀ = 16.5528 V
(b)
Maximum power
P₀ = E₀²/R...................... Equation 2
Where R = Resistance.
Given: R = 50 ohms, E₀ = 16.5528 V
Substitute into equation 2
P₀ = 50(16.5528)
P₀ = 827.64 W
