2 years ago

Cart 1 of mass m is traveling with speed 2vo in the +x-direction when it has an elastic collision with cart 2 of

Physics

1 answer:

iogann1982 [59]2 years ago

6 0

Answer:

Explanation:

Momentum conservation

$m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\$

Kinetic energy conservation

$\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2$

Solve the system

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A 8kg cat us running 4 m/s. How much kenetic energy is that

Flauer [41]

Using the Definition of Kinetic Energy, we have:

$E_{c}= \frac{mv^2}{2} \\ E_{c}= \frac{8*4^2}{2} \\ \boxed {E_{c}=64J}$

3 0

3 years ago

Si un vector tiene una dirección de 2300 a partir del eje x positivo, ¿Qué signos tendrán sus componentes x y y? Si la la razón

Masteriza [31]

If you can write this in English I can help

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2 years ago

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

3 0

3 years ago

What's the momentum (in kg-m/s) of a cannonball with a mass of 21 moving with a velocity of 25 m/s?

lbvjy [14]

P=mv
P=21*25=525 kg*m/s

8 0

3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

8 0

2 years ago