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DaniilM [7]
3 years ago
9

A simple pendulum, consisting of a mass on a string of length L, is undergoing small oscillations with amplitude A.

Physics
1 answer:
Leya [2.2K]3 years ago
4 0

Answer:

A. Period is halved

Explanation:

The period of a pendulum swing, T, is given in terms of mass as:

T = 2\pi \sqrt{\frac{I}{mgL} }

where I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity

h = Length of string

If the mass is increased by a factor of 4, that means:

M = 4m

(M = new mass)

The new period of the pendulum, T_n, will now be:

T_n = 2\pi \sqrt{\frac{I}{MgL} }\\\\\\T_n = 2\pi \sqrt{\frac{I}{4mgL} }\\\\\\T_n = 2\pi \sqrt{\frac{1}{4} * \frac{I}{mgL} }\\\\\\T_n = \frac{2\pi}{2}  \sqrt{\frac{I}{mgL} }\\\\\\T_n = \frac{1}{2} * 2\pi \sqrt{\frac{I}{mgL} }\\\\\\T_n = \frac{1}{2} * T

Hence, the period is halved.

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A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

a'=4\times \dfrac{v^2}{R}

a'=4\times a

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
Explain how characteristic and traits are related<br>​
jok3333 [9.3K]
Traits are basically your phenotype. They include things like hair color, height, and eye color. Alleles are versions of genes. ... This is a pretty basic idea of how traits and alleles are related.
4 0
3 years ago
The distance between two consecutive nodesof a standing wave is 20.9cm.Thehandgen-erating the pulses moves up and down throughac
irina1246 [14]

Answer:

Velocity, v = 0.239 m/s

Explanation:

Given that,

The distance between two consecutive nodes of a standing wave is 20.9 cm = 0.209 m

The hand generating the pulses moves up and down through a complete cycle 2.57 times every 4.47 s.

For a standing wave, the distance between two consecutive nodes is equal to half of the wavelength.

\dfrac{\lambda}{2}=0.209\ m\\\\\lambda=0.418\ m

Frequency is number of cycles per unit time.

f=\dfrac{2.57}{4.47}\\\\f=0.574\ Hz

Now we can find the velocity of the wave.

Velocity = frequency × wavelength

v = 0.574 × 0.418

v = 0.239 m/s

So, the velocity of the wave is 0.239 m/s.

4 0
3 years ago
At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
Over [174]

Answer:

v=32.9m/s

Explanation:

The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation a=v^2/r

This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:

v=\sqrt{ar} = \sqrt{12.5gr}=\sqrt{(12.5)(9.8m/s)(8.84m)}=32.9m/s

7 0
3 years ago
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
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