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mr_godi [17]
4 years ago
12

Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form

a charge of −7.50 nC? electrons (b) How many electrons must be removed from a neutral object to leave a net charge of 0.580 µC? electrons
Physics
1 answer:
Stels [109]4 years ago
8 0

Answer:

a) 4.681*10^10 electrons

b) 3.67*10^12 electrons

Explanation:

The amount of electrons in a charge of 1C is:

1C=6.2415*10^{18}\ electrons

You use the previous equality as a conversion factor.

a) The sing of the charge is not important in the calculation of the number electrons, so, you use the absolute value of the charge

7.50nC=7.50*10^{-9}C*\frac{6.2415*10^{18}}{1C}=4.681*10^{10}\ electrons

In 7.50nC there are 4.61*10^18 electrons

b)

0.580\mu C=0.580*10^{-6}C*\frac{6.2415*10^{18}}{1C}=3.67*10^{12}\ electrons

To obtain a charge of  0.580 µC in a neutral object you need to take out 3.67*10^12 electrons

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5 0
3 years ago
A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
loris [4]
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
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We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m

Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
      = </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
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Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
4 0
4 years ago
In 1967, new zealander burt munro set the world record for an indian motorcycle, on the bonneville salt flats in utah, with a ma
Fed [463]
Draw a velocity-time diagram as shown below.

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a = 26.82/4 = 6.705 m/s²
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The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
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Answer:
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The distance traveled during the acceleration phase is 502.6 m

3 0
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3 0
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The effect of gravity on a falling object can be modeled by a ball dropped
weeeeeb [17]

Answer:

B. Friction with air also affects the fall of the object.

Explanation:

The limitation of this experimental design is that friction with air also affects the fall of the object.

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