If a range hood removes contaminants at a flow rate of F liters of air per second, then the percent P of contaminants that are
1 answer:
Explanation:
The given value of P is as follows.
P = 1.06F + 22.18,
or, P' = 1.06
As p' is defined and non-zero. Hence, only critical points are boundary points.
For F = 10, the value of P will be calculated as follows.
P =
= 32.78
For F = 70, the value of P will be calculated as follows.
P =
= 96.38
Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.
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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:
proof in explanation
Explanation:
First, we will calculate the number of half-lives:
where,
n = no. of half-lives = ?
t = total time passed = 2100 million years
= half-life = 700 million years
Therefore,
Now, we will calculate the number of uranium nuclei left ():
and the rest of the uranium nuclei will become thorium nuclei ()
dividing both:
<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>