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3 years ago

How do the tension of the cord and the force of gravity affect a pendulum?

Physics

2 answers:

LuckyWell [14K]3 years ago

6 0

Answer:

Force of gravity may not affect a pendulum during its equilibrium state. But the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.

Similarly the tension in the cord will not affect the pendulum but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.

sleet_krkn [62]3 years ago

6 0

Answer:

A. They are the competing forces that act on a pendulum as it swings.

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You kick a soccer ball with a speed of 31 m/s at an angle of 50 degrees. How long does it take the ball to reach the top of its

Harlamova29_29 [7]

Answer:

Reaches max height at t = 2.42s.

Explanation:

I've assumed we are neglecting air resistance. If not let me know and I'll update.

We want to examine the behaviour of the ball in the y-direction. In the absence of air resistance the only force acting on the ball is gravity, which produces an acceleration in the negative y direction.

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3 years ago

A perturbation in the temperature of a stream leaving a chemical reactor follows a decaying sinusoidal variation, according to t

Vsevolod [243]

Derivating the function of the temperature and equating to 0, we can find the critical points:

$T'(t)=-a\cdot5e^{-at}\cdot\sin(wt)+w\cos(wt)\cdot5e^{-at}\\\\ 0=5e^{-at}(-a\sin(wt)+w\cos(wt))$

As $5e^{-at}\neq0$ :

$0=-a\sin(wt)+w\cos(wt)\iff a\sin(wt)=w\cos(wt)\iff \\\\\sin(wt)=\dfrac{w}{a}\cos(wt)\iff \tan(wt)=\dfrac{w}{a}\iff wt=\tan^{-1}\left(\dfrac{w}{a}\right)\iff \\\\\boxed{t=\dfrac{1}{w}\tan^{-1}\left(\dfrac{w}{a}\right)}$

Replacing:

$T(t)=5e^{-at}\cdot\sin(wt)=5e^{-\frac{a}{w}\tan^{-1}\left(\frac{w}{a}\right)}\cdot\sin(\tan^{-1}\left(\dfrac{w}{a}\right))$

We can reach: $\sin(\tan^{-1}\left(\dfrac{w}{a}\right))=\dfrac{w}{\sqrt{w^2+a^2}}$

Hence:

$T(t)=\dfrac{5w}{\sqrt{w^2+a^2}}e^{-\frac{a}{w}\tan^{-1}\left(\frac{w}{a}\right)}$

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3 years ago

Information about mars' moons.

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Answer:

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2 years ago

Why does the moon appear to wax grow larger and then wane or get smaller

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3 years ago

A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the v

seropon [69]

Answer:

v = 42.92 m/s

Explanation:

Given,

initial speed of the ball, v = 11 m/s

time taken to hit the ground = 5.5 m/s

velocity of the ball just before it hit the ground, v = ?

time taken by the ball to reach the maximum height

using equation of motion

v = u + at

final velocity = 0 m/s

0 = 11 - 9.8 t

t = 1.12 s.

time taken by the ball to reach the water from the maximum height

t' - 5.5 -1.12 = 4.38 s

using equation of motion for the calculation of speed just before it hit the water.

v = u + a t

v = 0 + 9.8 x 4.38

v = 42.92 m/s

Velocity of the ball just before it reaches the water is equal to v = 42.92 m/s

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3 years ago