A motor spins upward the flywheel with a persistent torque
of 50N⋅m.
What time does it take the flywheel to get to the top speed?
From the equation:
Tj = J*dω/dt
you can get the two equations:
Δt1= J1*Δω/Tj = 240*125.66/50 = 603.17 sec
Δt2= J2*Δω/Tj = 120*125.66/50 = 301.58 sec
Answer:
It would take 16.7 s for the work to be done by the engine.
Explanation:
From the question, given: Power = 390.3 kW
Work to be done = 6.5 x 
J
But, power and work done with respect to time, has a relationship of:
Power = 
So that,
time = 
Thus,
time = 
= 16.6539
time = 16.7 s
Time required is 16.7 seconds.
Thus, it would take 16.7 s for the work to be done by the engine.
1 is c
2 is false
3 false
4 i think is c ont 100% though
The net force on q2 will be 1.35 N
A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe force. Being a vector quantity, a force has both magnitude and direction.
Given Particles q1, q2, and q3 are in a straight line. Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C, and q3 = -2.50 x 10-6 C. Particles q₁ and q2 are separated by 0.500 m. Particles q2 and q3 are separated by 0.250 m.
We have to find the net force on q2
At first we will find Force due to q1
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²
F = 450 × 10⁻³
F₁ = 0.45 N (+)
Now we will find Force due to q2
F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²
F = 1800 × 10⁻³
F₂ = 1.8 N (-)
So net force (F) will be
F = F₂ - F₁
F = 1.8 - 0.45
F = 1.35 N
Hence the net force on q2 will be 1.35 N
Learn more about force here:
brainly.com/question/25573309
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