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3 years ago

Which has more thermal energy: lake or a cup of hot chocolate?

1 answer:

4 0

Though the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.

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What best describes desert pavements

Gekata [30.6K]

Answer:

Explanation:

A common theory suggests that desert pavements are formed through gradual removal of sand and other fine particles by the wind and intermittent rains leaving behind the large fragments. The larger rock particles are shaken into place by actions of different agents such as rain, wind, gravity, and animals

3 0

3 years ago

Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of move

svetoff [14.1K]

Answer:

I think its object 1

Explanation:

Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.

5 0

3 years ago

Help please it's urgent

insens350 [35]

I cant read it, i could most likely help if i could read it.

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3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

3 years ago

Why are the very first stars thought to have been much more massive than the sun? (a the clouds that made them were much more ma

pashok25 [27]

I think the correct answer would be that the temperature of the clouds that made the very first stars where thought to be higher since the clouds are made up of hydrogen and helium. Hope this answers the question. Have a nice day.

8 0

3 years ago

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