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2 years ago

A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

2 answers:

7 0

IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy = (mass) x   (gravity) x (height)

   = (0.15 kg) x (9.8 m/s²) x (20 m)

   = 29.4 kg-m²/s² = 29.4 joules .

mel-nik [20]2 years ago

3 0

The conservation of energy says that energy in an isolated system remains constant, that it is not created nor destroyed but transforms from one form of energy to another.
So when the ball is thrown into the air, it has kinetic energy (KE). As it goes into the air, that kinetic energy transforms into gravitational potential energy (GPE). The higher it gets, the more GPE it has and therefore the less KE it has.
At the top, all of it is transformed into GPE and there is no KE.
But total energy at any point after the ball is thrown until it lands it has the same energy throughout.

Knowing this, you can set the beginning KE equal to the final GPE.
KE = GPE

Insert formula for GPE.

KE = mgh
KE = 0.15kg*9.8m/s^2*20.0m
KE = 29 J

The ball initially had 29 J of kinetic energy

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3 years ago

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2 years ago

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The boiling point temperature of this substance when its pressure is 60 psia is 480.275 R

Explanation:

Given the data in the question;

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$(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}$

$(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }$

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Now,

$(\frac{dP}{dT} )_{sat } =$ $(\frac{P_2 - P_1}{T_2 - T_1})_{sat$

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we substitute;

$T_2$ $= ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}$

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2 years ago

You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

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Answer:

Explanation:

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$f(x,t)=Acos(kx-\omega t)$

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$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

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