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posledela
3 years ago
6

Please help brainliest ASAP...........

Physics
2 answers:
pantera1 [17]3 years ago
8 0

Answer:

I believe it is gender

Explanation:

ElenaW [278]3 years ago
6 0
I agree I think it’s gender.
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A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
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Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.
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2 years ago
A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra
son4ous [18]

Answer:

The answer is v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}.

Explanation:

From the law of gravity,

F = G \frac{Mm}{r^2}

considering F as a conservative force, F = - \nabla U,

the general expression for gravitational potential energy is

U = -G \frac{Mm}{r},

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

g = -G \frac{M}{r^2},

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

U = g M_e R,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

U_p = g M_e(R- r_s),

then, the kinetik energy when the rock hits the surface is

U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s),

so

v = \sqrt{2g(R-r_s)}

where g is the gravitational acceleration generated by the Sun at R,

g = G \frac{M_s}{R^2}.

8 0
3 years ago
A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac
zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

 ( because at moon g will become g/6)

As we know that

F=\dfrac{K}{x^2}

Here x= 1100 miles

F 2 tons

2=\dfrac{K}{1100^2}

So

K=2.4\times 10^6

We know that

Work = F. dx

W=\int_{x_1}^{x_2}F.dx

W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx

W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}

W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]

W=76.55 miles.metric tons

6 0
3 years ago
HELLO ONCE MORE FRIENDS . PLEASE HELP ME.
tatuchka [14]

Answer:

Do u want to talk?

Explanation:

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Which statement best describes the weather shown by the purple combination of semicircles and triangles on a line on a weather m
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Answer:

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Explanation:

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