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Sidana [21]
3 years ago
13

4) Block A has a mass of 3kg and velocity of 13m/s, catching up with a second block B that has a mass of 3kg and is moving with

5m/s in the same direction. They suffer an elastic collision and block B is now moving with 8m/s. What is the speed of block A after collision? 10m/s
8m/s
5m/s
13m/s
18m/s
Physics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

Option A is the correct answer.

Explanation:

Here momentum is conserved.

That is \left (m_Av_A+m_Bv_B \right )_{initial}=\left (m_Av_A+m_Bv_B \right )_{final}

Substituting values

    3\times 13+3\times 5=3v_A+3\times 8\\\\3v_A=39+15-24\\\\3v_A=30\\\\v_A=10m/s

Speed of block A after collision = 10 m/s

Option A is the correct answer.

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Answer:

Explanation:

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3 0
3 years ago
A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi
Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

         \Delta E_{system} = E_{in} - E_{out}

        \Delta U = W_{in} - Q_{out}

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Therefore, we will calculate the final temperature as follows.

            \frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}

       T_{2} = \frac{20 psia}{14.7}(638 R)

                   = 868.03 R

Now, we will calculate the mass as follows.

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                 = \frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}

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Hence,

        W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Putting the values into the above equation as follows.

            W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

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6 0
3 years ago
1. A 180 kg motorcycle travels in a straight line on a horizontal road. The relationship between
Sergio039 [100]
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5 0
3 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

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Direction: negative

Explanation:

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ma = qVB

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a = -0.33 m/s² k^

7 0
3 years ago
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4 0
3 years ago
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