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3 years ago

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You are on a new planet, and are trying to use a simple pendulum to find the gravity experienced on this new planet. You find th

at a pendulum having a length of 35 cm has a period of 2 seconds. What is the acceleration due to gravity on this planet?
A. 13.82 m/s^2
B. 345.4 m/s^2
C. 3.45 m/s^2
D. 1.1 m/s^2

1 answer:

levacccp [35]3 years ago

4 0

To solve this problem it is necessary to apply the concepts related to the Period based on the length of its rope and gravity, mathematically it can be expressed as

$T= 2\pi \sqrt{\frac{L}{g}}$

g = Gravity

L = Length

T = Period

Re-arrange to find the gravity we have

$g = \frac{4\pi^2 L}{T^2}$

Our values are given as

$L = 0.35m\\T = 2s\\$

Replacing we have

$g = \frac{4\pi^2 L}{T^2}$

$g = \frac{4\pi^2 0.35}{2^2}$

$g = 3.45 m/s^2$

Therefore the correct answer is C.

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Answer:

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When does a star reach equilibrium?

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3 years ago

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

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Answer:

Stay the same

Explanation:

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From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

$V= \frac{Q}{C}$

where

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we said that:

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therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

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3 years ago

Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3

cupoosta [38]

Answer:

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Where $\frac{\Delta c}{\Delta x}$ is the concentration of gradient

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and J is the flux of atoms.

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Where $D_0$ is the proportionality constant,

R is the gas constant, T the temperature and $E_d$ is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

$J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}$

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$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

We have all the values in our equation.

$\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}$

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$J = 3.2*10^{-9}m^2/s$

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Substituting,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

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3 years ago