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Bezzdna [24]
3 years ago
5

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Physics
1 answer:
Vlad1618 [11]3 years ago
5 0

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

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

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

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What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An
GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}

E=\frac{2m_{0}c^}2}{\sqrt{3}}

E=\frac{2\times0.001\times9\times10^{16}}{1.732}

E=1.04\times10^{14}J

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV

4 0
3 years ago
An object with a mass of 10 kg is accelerated upward at 2 m/sec2. What force is required?
telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

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Answer:

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Explanation:

7 0
3 years ago
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