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xeze [42]
4 years ago
14

Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line towar

d each other with velocities +2.00 meters/second and -1.30 meters/second respectively. What is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?
Physics
2 answers:
cestrela7 [59]4 years ago
8 0
Two things:
1) All collisions conserve momentum
2) Elastic collisions also conserve kinetic energy.

So basically the kinetic energy before the collision is the same as the kinetic energy after the collision.

KE = ½mv²

For object A:
KE = ½(7.20 kg)(2.00 m/s)² = 14.4 J

For object B:
KE = ½(5.75 kg)(-1.30 m/s)² = 4.86 J

total kinetic energy = 14.4 J + 4.86 J = 19.3 J
Genrish500 [490]4 years ago
7 0
The equation for Kinetic Energy in an Elastic Collision is this:
1/2(m₁v₁) + 1/2(m₂v₂) = 1/2(7.20kg)(2.00m/s) + 1/2(5.75kg)(-1.30m/s) =
3.4625m/s

I think....technically the equation above equates to the same thing except measuring the final velocity of the given masses, however you were not given that in the equation, making the first half of the full equation the only viable solution.

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During free fusion, the eyes _______ in order to view a stereogram without a stereoscope. A. use the pictorial depth cue B. are
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Answer:

D

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7 0
3 years ago
A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of
nexus9112 [7]

Answer:

F_1=366.67Hz

Explanation:

From the question we are told that:

Frequency of 3rd harmonics F_3=1100

Frequency of 5th harmonics F_3=1833

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

 \lambda_3=\frac{4}{3}l

Therefore

 F_3=\frac{3v}{4l}

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

 \lambda_1=\frac{4}{1}l

Therefore

 F_1=\frac{v}{4l}

Generally the equation for the frequency of the first harmonic is mathematically given by

 F_1=\frac{F_3}{3}

 F_1=\frac{1100}{3}

 F_1=366.67Hz

7 0
3 years ago
a train engineer started the train from a standstill and sped up to 5 meters per second, she then rounded a corner at a constant
olasank [31]

for acceleration we can define that rate of change in velocity is know as acceleration

So whenever velocity of train is changing with time we can say train is accelerating

Now here if initially train is standstill then after some time its speed is 5 m/s

so here the train is accelerated first time

Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time

After this train chugged around a curve with same speed 10 m/s

SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration

SO this is accelerated Third time

Then its speed decreases and it comes to speed of 5 m/s from 10 m/s

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Then finally train comes to stop so again its speed changed from 5 m/s to 0

so this is acceleration of train Fifth time

So total train will accelerate 5 times in whole path

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3 years ago
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3 0
3 years ago
Question 10(Multiple Choice Worth 2 points)
Svetach [21]
P = mv

P momentum
m mass
v velocity
3 0
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