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mylen [45]
2 years ago
12

A 2.5g copper penny is given a charge of -4.0*10^-9c. how mny excess electrons are on the penny?

Physics
1 answer:
Lyrx [107]2 years ago
7 0

Answer : The excess of electrons on the penny are, 2.5\times 10^{10} electrons

Solution : Given,

Total charge = -4.0\times 10^{-9}C

Charge on electron = -1.6\times 10^{-19}C

Formula used :

\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}

Now put all the given values in this formula, we get the excess of electrons present on the penny.

\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}=\frac{-4.0\times 10^{-9}C}{-1.6\times 10^{-19}C/e^-}=2.5\times 10^{10}e^-

Therefore, the excess of electrons on the penny are, 2.5\times 10^{10} electrons


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A box of paper is labeled 24lb paper. This means that 500 sheets (counted number) of paper size 17in x 22in weighs 24 pounds (Th
Tresset [83]

Answer:

The mass of a single paper  is approximately 0.047 lb/paper which in SI Units is approximately 21.77  g/paper

Explanation:

The given information on the size and the weight of paper are;

The mass of a box of 500 sheets of paper = 24 lb

The number of sheets in the paper = 500 sheets

The dimensions of the paper = 17 in. × 22 in., which is equivalent to  43.18 cm × 55.88 cm

The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)

The mass of a single paper = 24 lb/500 = 0.047 lb/paper

Given that 1 lb = 453.6 g, we have;

0.047 lb/paper = 0.047 lb/paper×453.6 g/(lb) = 21.77  g/paper

The mass of a single paper  = 0.047 lb/paper = 21.77  g/paper.

6 0
3 years ago
Select the correct answer.
mamaluj [8]

Answer:

amount of charge on the source charge

6 0
2 years ago
A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when
vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}

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5 0
3 years ago
A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

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2 years ago
Ozone gas absorbs the ultraviolet radiations of the sun in the
bija089 [108]
Hello,

The answer is option C "<span>thermosphere".</span><span>

Reason:

The troposphere is the layer of the air that the harmful heat rays (from the sun is mostly absorbed). The thermosphere is directly above the exosphere which is the sphere with the thin layer of the sphere therefore the answer should be option C.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>
7 0
3 years ago
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