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Kisachek [45]
3 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]3 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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What is the electrical force between q2 and q3? Recall that k = 8. 99 × 109 N•meters squared over Coulombs squared. 1. 0 × 1011
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Force on the particle is defined as the application of the force field of one particle on another particle. the electrical force between q₁ and q₃ will be –1. 1 × 10¹¹ N.

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Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The electric force in the second case will be the same as in the first case. Therefore the force on the particle will be the same.

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

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