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Kisachek [45]
2 years ago
13

Select the correct answer. In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the vel

ocity of the ball changed to -0.4 meter/second. The collision takes 0.2 seconds to occur. What's the acceleration of the ball during the collision? Use a equals fraction numerator v minus u over denominator t end fraction. Group of answer choices 5.0 m/s2 -5.0 m/s2 1.0 m/s2 0.24 m/s2 -1.0 m/s2
Physics
1 answer:
JulijaS [17]2 years ago
4 0

Answer:

a = - 5 m/s²

Explanation:

The acceleration of the ball can be found by using the general formula of acceleration as follows:

a = \frac{\Delta V}{\Delta t}

where,

a = acceleration of ball = ?

ΔV = change in velocity = Final Velocity - Initial Velocity

ΔV = - 0.4 m/s - 0.6 m/s = -1 m/s

Δt = time of contact with the wall = 0.2 s

Therefore, using the values in equation, we get:

a = \frac{-1\ m/s}{0.2\ s}\\\\a = -5\ m/s^2

Therefore, the correct option is:

<u>a = - 5 m/s²</u>

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Read 2 more answers
A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What
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Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

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We need to calculate the kinetic energy of the anti proton

Using formula of energy

E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}

We know that,

m_{p}c^2=m_{np}c^2

So, E_{photon}=2mc^2+K.E_{p}+K.E_{np}

K.E_{np}=E_{photon}-(2mc^2+K.E_{p})

Put the value into the formula

K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}

K.E_{np}=147.4\ MeV

Hence, The kinetic energy of the anti proton is 147.4 MeV.

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