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2 years ago

What is the period of a 3. 98 m long pendulum? What is the period of a 99. 4 cm long pendulum?.

1 answer:

4 0

A pendulum's time period is the amount of time it takes to complete one oscillation. The pendulum with a length of 3.98 m and a length of 99.4 cm will have a time period of 4 and 1.99 seconds, respectively.

<h3>What is the definition of a simple pendulum?</h3>

A simple pendulum is a device with a point mass hanging from fixed support and connected to a light inextensible thread.

The vertical line going through the fixed support represents the mean position of a basic pendulum.

h = vertical distance from the point of suspension

L = length of the simple pendulum

The time period of the simple penduum is given by

$\rm T = 2\pi\sqrt{\frac{l}{g} }$

For l = 3.98 m

$\rm T = 2\times 3.14 \sqrt{\frac{3.98}{9.81} }$

T = 4 second

For l = 99.4 cm = 0.994 m

$\rm T = 2\times 3.14 \sqrt{\frac{0.994}{9.81} }$

T = 1.99 second

The time period of the pendulum having 3.98 m and 99.4 cm in length will be 4 and 1.99 seconds respectively.

To learn more about the simple pendulum refers to the link;

brainly.com/question/14759840

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3 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

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Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

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3 years ago

Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

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Explanation:

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