The question doesn't describe any experiment. If the same experiment is repeated, no matter how many times, the acceleration due to gravity will remain the same as it was during the non-existent original experiment, and will have no effect on anything.
A ammeter. I hope this helps.
Answer:
26.17 V
Explanation:
In a alternator , an ac voltage is produced whose magnitude of voltage varies sinusoidally.
Maximum voltage induced = nBAω
where n is no of turns , B is magnetic field , A is area of the coil and ω is angular velocity .
Rate of rotation n = 1000rpm = 1000/60 rps = 16.67 rps
angular velocity ω = 2π n = 2π x 16.67 = 104.68 rad / s
Putting all values in the given equation above
Max voltage = 250 x .1 x .01 x 104.68
= 26.17 V
Answer:
b. 1.9 Ω
Explanation:
Here is the complete question
A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? a. 1.5 ? b. 1.9 ? c. 5.0 ? d. 14 ?
Solution
The voltage across the inductor V is
where V₀ = emf of source = 14 V, R = resistance, L = inductance = 34 H and t = time
Given that V = 80% of its maximum value after 4.0 s, this implies that V = 80 % of V₀ = 0.8V₀ and t = 4.0 s
Since
and V = 0.8V₀.
Since we need to find R, we make R subject of the formula, we have
![V = V_{0}e^{-\frac{Rt}{L} }](https://tex.z-dn.net/?f=V%20%3D%20V_%7B0%7De%5E%7B-%5Cfrac%7BRt%7D%7BL%7D%20%7D)
![V/V_{0}= e^{-\frac{Rt}{L} }](https://tex.z-dn.net/?f=V%2FV_%7B0%7D%3D%20e%5E%7B-%5Cfrac%7BRt%7D%7BL%7D%20%7D)
taking natural logarithm of both sides, we have
㏑(V/V₀) = -Rt/L
R = -L㏑(V/V₀)/t
Substituting the values of the variables into the equation, we have
R = -34㏑(0.8V₀/V₀)/4.0 s
R = -34㏑(0.8)/4.0 s
R = -34 × -0.2231/4.0 s
R = 7.587/4
R = 1.896 Ω
R ≅ 1.9 Ω
So, B is the answer