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2 years ago

What is the kinetic energy of 0.01kg bullet traveling at a velocity of 700m/s ?

Physics

1 answer:

Nata [24]2 years ago

5 0

Answer:

kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie.

Explanation:

In the given question, we are told that what is the kinetic energy of mass M equals 0.1 kg bullet traveling at a velocity Velocity is given and 700 m/s. So we know that kinetic energy mm-hmm k equals one half m v squared. So this will be mass is given 0.1 and velocity is 700 so 700 square this is one half 0.1 in two 49 double zero, double zero. This is one-half into 49 double zero. So kinetic energy is 2, 4, 50 thirties, 2.45 into 10, raise to three Julie. This is kinetic energy. Thank you.

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3 years ago

A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k

Veronika [31]

Answer:

$14.98\ \text{kg m/s}$

$45.26^{\circ}$

Explanation:

$P_1$ = Initial momentum of the pin = 13 kg m/s

$P_i$ = Initial momentum of the ball = 18 kg m/s

$P_2$ = Momentum of the ball after hit

$55^{\circ}$ = Angle ball makes with the horizontal after hitting the pin

$\theta$ = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

$P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}$

Momentum in the y direction

$P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}$

$(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}$

The pin's resultant velocity is $14.98\ \text{kg m/s}$

$P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}$

The pin's resultant direction is $45.26^{\circ}$ below the horizontal or to the right.

4 0

3 years ago