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larisa86 [58]
2 years ago
15

A 2.0-kg ball rolls to the right at 3.0 m/s. A 4.0-kg ball rolls to the left at 2.0 m/s . What is the momentum of the system aft

er a head-on collision of the two balls?
Physics
1 answer:
charle [14.2K]2 years ago
4 0

Answer:

Final momentum after a head on collision is -2kgm/s

Explanation:

         One ball moves to the right and the other moves opposite  and momentum is a vector quantity so that considering the direction

Initial momenta are        P₁=2x3=6kgm/s        P₂=4x(-2)=-8kgm/s      

Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s

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Layered means that the chemicals are not soluble in one another. B is not the answer.

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D a solution is not a pure substance (singular) by itself. There are at least 2 chemicals together.

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Which characteristic should a good scientific question have?
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A large truck collides with a small car. True or False: The truck exerted a greater magnitude force on the car than the car exer
Reptile [31]

Answer:

False.

Explanation:

The forces on the car and truck are equal and opposite. The equal forces cause accelerations of the truck and car inversely proportional to their mass. That is, If the Truck A exerts a force FAB on car B, then the car will exert a force FBA on the truck. Therefore,

FBA = −FAB

However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:

atruck = F/M

acar = F/m

and combining these we get:

atruck/acar = m/M

So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

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d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

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