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Oksanka [162]
3 years ago
9

A car travels 2155.0m in 195.9s. What is the car's average speed?

Physics
1 answer:
inna [77]3 years ago
3 0

Average speed of the car is 11 m/s

Explanation:

  • Speed is calculated by the rate of change of displacement.
  • It is given by the formula, Speed = Distance/Time
  • Here, distance = 2155 m and time = 195.9 s

Speed of the car = 2155/195.9 = 11 m/s

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A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

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9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

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=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

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Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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Answer:

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Explanation:

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Where k=spring constant

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h=2m

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e=20cm to meter 20/100= 0.2m

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First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

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