Answer:
Explanation:
Given
Free fall acceleration on mars ![g_{m}=3.7\ m/s^2](https://tex.z-dn.net/?f=g_%7Bm%7D%3D3.7%5C%20m%2Fs%5E2)
Time Period of pendulum on earth ![T=1\ s](https://tex.z-dn.net/?f=T%3D1%5C%20s)
Time period of Pendulum is given by
![T=2\pi \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
for earth
![1=2\pi\cdot \sqrt{\frac{L}{9.8}}](https://tex.z-dn.net/?f=1%3D2%5Cpi%5Ccdot%20%5Csqrt%7B%5Cfrac%7BL%7D%7B9.8%7D%7D)
![L=\frac{9.8}{(2\pi )^2}](https://tex.z-dn.net/?f=L%3D%5Cfrac%7B9.8%7D%7B%282%5Cpi%20%29%5E2%7D)
![L=0.498\ m](https://tex.z-dn.net/?f=L%3D0.498%5C%20m)
(b)For same time period on mars length is given by
![L'=\frac{g_m}{(2\pi )^2}](https://tex.z-dn.net/?f=L%27%3D%5Cfrac%7Bg_m%7D%7B%282%5Cpi%20%29%5E2%7D)
![L'=\frac{3.7}{39.48}](https://tex.z-dn.net/?f=L%27%3D%5Cfrac%7B3.7%7D%7B39.48%7D)
![L'=0.0936\ m](https://tex.z-dn.net/?f=L%27%3D0.0936%5C%20m)
The period of the pendulum is 8.2 s
Explanation:
The period of a simple pendulum is given by the equation:
![T=2\pi \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
where
L is the length of the pendulum
g is the acceleration of gravity
T is the period
We notice that the period of a pendulum does not depend at all on its mass, but only on its length.
For the pendulum in this problem, we have
L = 16.8 m
and
(acceleration of gravity)
Therefore the period of this pendulum is
![T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B16.8%7D%7B9.8%7D%7D%3D8.2%20s)
#LearnWithBrainly
Answer:
Explanation:
Please check the picture and consider straight lines
Answer:
R = 2481 Ω
L= 1.67 H
Explanation:
(a) We have an inductor L which has an internal resistance of R. The inductor is connected to a battery with an emf of E = 12.0 V. So this circuit is equivalent to a simple RL circuit. It is given that the current is 4.86 mA at 0.725 ms after the connection is completed and is 6.45 mA after a long time. First we need to find the resistance of the inductor. The current flowing in an RL circuit is given by
i = E/R(1 -e^(-R/L)*t) (1)
at t --> ∞ the current is the maximum, that is,
i_max = E/R
solve for R and substitute to get,
R= E/i_max
R = 2481 Ω
(b) To find the inductance we will use i(t = 0.940 ms) = 4.86 mA, solve (1) for L as,
Rt/L = - In (1 - i/i_max
)
Or,
L = - Rt/In (1 - i/i_max
)
substitute with the givens to get,
L = -(2481 Si) (9.40 x 10-4 s)/ In (1 - 4.86/6.45
)
L= 1.67 H
<u><em>note :</em></u>
<u><em>error maybe in calculation but method is correct</em></u>
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