The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.
Answer:
The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Explanation:
Given:
(A)
Current A
Voltage V
For finding the resistance,
12Ω
(B)
For finding power delivered,
Watt
(C)
For finding the potential difference,
V
Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Answer:
F' = (3/2)F
Explanation:
the formula for the electric field strength is given as follows:
E = F/q
where,
E = Electric Field Strength
F = Force due to the electric field
q = magnitude of charge experiencing the force
Therefore,
F = E q ---------------- equation (1)
Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:
F' = (E/2)(3 q)
F' = (3/2)(E q)
using equation (1)
<u>F' = (3/2)F</u>
Answer:
The number of bright fringes per unit width on the screen is,
Explanation:
If d is the separation between slits, D is the distance between the slit and the screen and is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :
is the wavelength
n is the order
If n = 1,
So, the the number of bright fringes per unit width on the screen is . Hence, the correct option is (B).