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horsena [70]
3 years ago
8

A pan of ice is placed over a flame.

Physics
1 answer:
AveGali [126]3 years ago
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 .

3 0
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

C'=\frac{\epsilon_0 A}{d'}=\frac{\epsilon_0 A}{2d}=\frac{C}{2}

And therefore, the new charge stored on the capacitor will be

Q'=C'V=\frac{C}{2}V=\frac{Q}{2}

6 0
3 years ago
Find the Gravitational Potential at a point on the earth’s surface. Take mass of earth as 5.98 X 10 24 kg, its radius as 6.38 X
kherson [118]

Answer:

-6.25 x 10^7 J/Kg\\\\

Explanation:

The final answer is  -6.25 x 10^7 J/Kg\\\\

5 0
3 years ago
The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reac
Anna35 [415]

Answer:

The correct answer is "53.15 days".

Explanation:

Given that:

Half life of 131_{I},

T_{\frac{1}{2} }= 8 \ days

  • Let the initial activity be "R_o".
  • and, activity to time t be "R".

To find t when R will be "1%" of R_o, then

⇒ R=\frac{1}{100}R_o

As we know,

⇒ R=R_o e^{-\lambda t}

or,

∴ e^{\lambda t}=\frac{R_o}{R}

By putting the values, we get

        =\frac{R_o}{\frac{R}{100} }

        =100

We know that,

Decay constant, \lambda = \frac{ln2}{T_{\frac{1}{2} }}

hence,

⇒ \lambda t=ln100

     t=\frac{ln100}{\lambda}

        =\frac{ln100}{\frac{ln2}{8} }

        =53.15 \ days  

5 0
3 years ago
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