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Andrews [41]
4 years ago
5

A boy and his younger sister are at the zoo on a hot day. They each buy a cold lemonade. The boy buys a large lemonade, and his

sister buys a small. They set their cups on a hot table and wait a while before they start drinking. When they finally begin to drink their lemonade, the girl complains that her drink is no longer cold. However, the brother states that his larger drink is still cold.
What conclusions could the boy and girl accurately draw from their observations? Choose the two that apply.

Question options:

A If there are two different amounts of the same liquid, the temperature is a good way to compare their thermal energies.


B The particles in the warmer cup of lemonade have more kinetic energy than the particles in the colder cup of lemonade.


C Energy from the hot table and warm air is transferred to the cold cups of lemonade.


D The thermal energy of a liquid is the same regardless of
Physics
1 answer:
FromTheMoon [43]4 years ago
7 0

Answer: A and C (i took the test)

Explanation: Hope this helps:)

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1)the magnetic field is strongest near_________of a bar magnet.
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3 years ago
There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
taurus [48]

Answer:

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Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

The electric potential at a distance d due to 'Q' is:

V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

V_{total} = \frac{q}{4\pi\epsilon_{o}d}

Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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Answer:

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