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Phoenix [80]
3 years ago
14

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

n the same direction at vTi 5 20.0 m/s (Fig. P9.22). The velocity of the car immediately after the collision is vCf 5 18.0 m/s to the east. (a) What is the velocity of the truck immediately after the colli
Physics
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it.  Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s  ==>  (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out.  If you prefer to use the formula,
the first problem you run into is:  You need to remember the formula !

The formula is        D = 1/2 a T²

                   Distance = (1/2 acceleration) x (time in seconds)²

             Acceleration = (change in speed) / (time for the change)
                                  =        (18 m/s)            /        (12 sec)
                                  =                      1.5 m/s² .

                  Distance  =  (1/2 x 1.5 m/s²) x (12 sec)²
                                  =       (0.75 m/s²)  x  (144 sec²)  =  108 meters .

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3 years ago
Which is true of the greenhouse effect?
Aleksandr [31]

Answer:

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3 0
3 years ago
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

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