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aalyn [17]
3 years ago
15

A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?

Physics
2 answers:
Wewaii [24]3 years ago
6 0
75 km/h is the speed of the car

Akimi4 [234]3 years ago
5 0
.02 km/s
Hope this helped
(distance divided by time= speed)
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A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is
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Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

m_1 = Mass of block = 0.072 kg

m_2 = Mass of bullet = 4.67 g

u_1 = Initial Velocity of block = 0

u_2 = Initial Velocity of bullet = 629 m/s

v_1 = Final Velocity of block = 17 m/s

v_2 = Final Velocity of bullet

In this system the linear momentum is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J

The final kinetic energy

K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J

Initial Kinetic energy > Final kinetic energy

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