Answer:
α = - 1.883 rev/min²
Explanation:
Given
ωin = 113 rev/min
ωfin = 0 rev/min
t = 1.0 h = 60 min
α = ?
we can use the following equation
ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t
⇒ α = (0 rev/min - 113 rev/min) / (60 min)
⇒ α = - 1.883 rev/min²
Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a).
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
<span>Radius = 150 cramwells </span>
<span>t = sqrt(150/1.6) = 9.68 s.
I hope this helps.</span>
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:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
Answer:
takes 365 days and it bring us seasons such as, spring,winter and fall.
