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3 years ago

What is the free-fall acceleration on a planet where the period of a 1.07 m long pendulum is 2.02 s?

Physics

1 answer:

sergeinik [125]3 years ago

5 0

Answer:

10.34 m/s^2

Explanation:

length, L = 1.07 m

Time period, T = 2.02 s

The formula of the time period of the pendulum is

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

where, g is the free fall acceleration on that planet.

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

$g=4\pi ^{2}\frac{1.07}{2.02^{2}}$

g = 10.34 m/s^2

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Answer:

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Explanation:

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4 years ago

By what factor does the peak frequency change if the celsius temperature of an object is doubled from 20.0 ∘c to 40.0 ∘c?

mart [117]

Answer:

it increases by a factor 1.07

Explanation:

The peak wavelength of an object is given by Wien's displacement law:

$\lambda=\frac{b}{T}$ (1)

where

b is the Wien's displacement constant

T is the temperature (in Kelvins) of the object

given the relationship between frequency and wavelength of an electromagnetic wave:

$f=\frac{c}{\lambda}$

where c is the speed of light, we can rewrite (1) as

$\frac{c}{f}=\frac{b}{T}\\f=\frac{Tc}{b}$

So the peak frequency is directly proportional to the temperature in Kelvin.

In this problem, the temperature of the object changes from

$T_1 = 20.0^{\circ}+273=293 K$

$T_2 = 40.0^{\circ}+273 = 313 K$

so the peak frequency changes by a factor

$\frac{f_2}{f_1} \propto \frac{T_2}{T_1}=\frac{313 K}{293 K}=1.07$

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Answer:

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