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Anastasy [175]
3 years ago
15

A 1120 kg car traveling initially with a speed of 27.40 m/s in an easterly direction crashes into the back of a 9 490 kg truck m

oving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east. (a) What is the velocity of the truck right after the collision?
Physics
1 answer:
asambeis [7]3 years ago
8 0

Answer:

41.5 m/s

Explanation:

Since there are no external forces, momentum must be conserved.

Momentum P = mv.

Total momentum before the collision:

P = m_{car}v_{car} + m_{tru}_{ck}v_{tru}_{ck} = 1120 * 27.4 + 490 * 20 = 40.488

Total momentum after the collision:

P = 1120 * 18 + 490 * v_{tru}_{ck} = 40.488

solving for v:

v_{truck} = 41.5

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Consider two soap bubbles with radius r1 and r2 (r1 <r2) connected via a valve. What happens if we open the valve​
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Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

3 0
3 years ago
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