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SSSSS [86.1K]
3 years ago
8

A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th

e rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
Physics
1 answer:
zhannawk [14.2K]3 years ago
3 0

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

        v = ?

        u = 0 m/s

        a = 4.0 m/s²

         t = 5 s

         v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

        v = ?

        u = 20 m/s

        a = 0 m/s²

         t = 10 s

         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

        v = ?

        u = 20 m/s

        a = -2.0 m/s²

         t = 4 s

         v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

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Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o
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Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, Q_1 and Q_2, separated by a distance, d, is given

F = k\dfrac{Q_1Q_2}{r^2}

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

F_X = k\dfrac{18Q}{x^2}

Force due to charge Y

F_Y = k\dfrac{-27Q}{(0.60-x)^2}

Since both forces are equal and opposite,

F_X = -F_Y

k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}

\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}

2(0.60-x)^2 = 3x^2

2(0.36-1.20x+x^2) = 3x^2

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x^2+2.40x-0.72 = 0

Applying the quadratic formula,

x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}

x = -2.7 or x = 0.27

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0
3 years ago
A 0.72-m long string has a mass of 4.2 g. The string is under a tension of 84.1 N. What is the speed of a wave on this string?
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The speed of the wave in the string is 83.4 m/s

Explanation:

For a standing wave in a string, the speed of the wave is given by the equation:

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In this problem, we have:

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Solving the equation, we find the speed of the wave:

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Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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