Answer:
the net toque is τ=8.03* 10⁻⁴ N*m
Explanation:
Assuming the disk has constant density ρ, the moment of inertia I of is
I = ∫r² dm
since m = ρ*V = ρπR² h , then dm= 2ρπh r dr
thus
I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2
replacing values
I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²
from Newton's second law applied to rotational motion
τ= Iα , where τ=net torque and α= angular acceleration
since the angular velocity ω is related with the angular acceleration through
ω= ωo + α*t → α =(ω-ωo)/t = (21 rad/s-0)/0.8 s = 26.25 rad/s²
therefore
τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m
The first one is electrical energy
Explanation:
PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA
= 65 kg(
) = 637 N
CON LA FÓRMULA DE LA ENERGÍA POTENCIAL
U = 637 N(6000 m - 3500 m) = 1592500 J
Answer:
B) Diphosphorus pentoxide
Explanation:
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.
