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Darina [25.2K]
3 years ago
14

A pendulum has 201 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o

f its swing?
Physics
1 answer:
malfutka [58]3 years ago
7 0
<h2>Hello!</h2>

The answer is: 201 J of kinetic energy.

<h2>Why?</h2>

The motion of a pendulum (with no friction considered) is a continuous exchange between potential energy and kinetic energy.

So, at the highest point of its swing, the potential energy will be the maximum potential energy that the pendulum can have and the kinetic energy will be 0 since at max height the speed tends to 0.

On the opposite side, when the pendulum is at the bottom (the lowest point of its swing) the potential energy will be the minimum (tends to 0) but the kinetic energy will be the maximum.

Also, in the pendulum motion, the total energy is conserved, meaning that:

PE_{h} +KE_{h}=PE_{l} +KE_{l}\\PE=mgh\\KE=\frac{1mv^{2} }{2}

Where,

PE(h),is the potential energy at the highest point.

KE(h), is the kinetic energy at the highest point.

PE(l), is the potential energy at the lowest point (bottom of pendulum swing).

KE(l), is the kinetic energy at the lowest point (bottom of pendulum swing).

m, is the mass of the object.

g, is the acceleration of gravity.

v, is the speed of the object.

So, what is the energy at the bottom of its swing?

201J +0=0 +201J\\201J=201J

So, the pendulum has 201 of kinetic energy at the bottom of its swing.

Meaning that the energy is conserved.

Have a nice day!

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<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed  = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

 

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

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impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

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3 years ago
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

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v(f) = 0.75 / 0.05

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