A pan of ice is placed over a flame.

1 answer:

8 0

A. The molecules start packed together very tightly in a solid. Then when it turns to water, the molecules can move around each other freely, but still contained. When water turns to vapor, the molecules are going crazy moving around. They are not contained at all and bounce of of each other freely.

b. The temperature rises. (ice turns to water at 33 degrees and water turns to vapor at 212 degrees)

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If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

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3 years ago

Convert the number from scientific into standard notation: 5.9 x 10-2

guapka [62]

Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)

6 0

3 years ago

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant

Natasha2012 [34]

Answer:

D) Q/2

Explanation:

The relationship between charge Q, capacitance C and voltage drop V across a capacitor is

$Q=CV$ (1)

In the first part of the problem, we have that the charge stored on the capacitor is Q, when the voltage supplied is V. The capacitance of the parallel-plate capacitor is given by

$C=\frac{\epsilon_0 A}{d}$

where $\epsilon_0$ is the vacuum permittivity, A is the area of the plates, d is the separation between the plates.

Later, the voltage of the battery is kept constant, V, while the separation between the plates of the capacitor is doubled: $d'=2d$ . The capacitance becomes