Headlights must be activated during times of darkness and when visibility is limited to what ?
1 answer:
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The emf induced in the second coil is given by:
V = -M(di/dt)
V = emf, M = mutual indutance, di/dt = change of current in the first coil over time
The current in the first coil is given by:
i = i₀
i₀ = 5.0A, a = 2.0×10³s⁻¹
i = 5.0e^(-2.0×10³t)
Calculate di/dt by differentiating i with respect to t.
di/dt = -1.0×10⁴e^(-2.0×10³t)
Calculate a general formula for V. Givens:
M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)
Plug in and solve for V:
V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))
V = 320e^(-2.0×10³t)
We want to find the induced emf right after the current starts to decay. Plug in t = 0s:
V = 320e^(-2.0×10³(0))
V = 320e^0
V = 320 volts
We want to find the induced emf at t = 1.0×10⁻³s:
V = 320e^(-2.0×10³(1.0×10⁻³))
V = 43 volts
Answer:
<em>The second option has a lower power output. P=30 W</em>
Explanation:
<u>Mechanical Power </u>
It is a physical magnitude that measures the rate a work W is done over time t.
Since W=F.d
The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of
The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is
The second option has a lower power output
Explanation:
We have,
Mass of an object is 0.5 kg
Force constant of the spring is 157 N/m
The object is released from rest when the spring is compressed 0.19 m.
(A) The force acting on the object is given by :
F = kx
(B) The force is simply given by :
F = ma
a is acceleration at that instant