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2 years ago

6

A 2kg toy car traveling at 2m/s strikes a second toy car at rest. The mass of the second toy car is 3. The two stock together an

d move off together. What is their resultant speed?

1 answer:

Nutka1998 [239]2 years ago

7 0

Mass of first car = Initial mass (Mi) = 2 kg
Initial velocity (Vi) = 2 m/s

Mass of both cars together = Final mass (Mf) = 2 + 3 kg = 5 kg
Final Velocity (Vf) = ?

Applying law of conservation of momentum,

Mi x Vi = Mf x Vf
2 x 2 = 5 x Vf
Vf = 4/5 = 0.8 m/s

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A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

3 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

$s_1 = v_1t = 1.6t$

Similarly the distance (m) from your friend building to the restaurant:

$s_2 = v_2t = 1.2t$

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

$s_1 + s_2 = 5b$

$1.6t + 1.2t = 5b$

$2.8t = 5b$

$t = 5b/2.8 = 25b/14$

$s_2 = 1.2t = 1.2(25b/14) = 2.143b$

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

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3 years ago

Weight of the body will be more in air water Milk or Mercury

Montano1993 [528]

Answer:

Mercury is the answer..

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2 years ago

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The inner planets or call terrestrial because they

AnnyKZ [126]

They are similar to the earth. Not relative to the outer planets

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3 years ago