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kupik [55]
3 years ago
6

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

etween them has a magnitude of 0.235 N.
What is the charge on each sphere if the two charges are equal?
What is the charge on first sphere if it has four times the charge of the other?
What is the charge on the second sphere?
Physics
1 answer:
sdas [7]3 years ago
6 0

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}

q=7.56\times10^{-7}C

Thus, the charge on each sphere is q=7.56\times10^{-7}C.

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}

q=3.78\times10^{-7}C

Thus, the charge on second sphere is q=3.78\times10^{-7}C and the charge on first sphere is 4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C.

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Answer:

1.07\cdot 10^{-3} m

Explanation:

The peak wavelength of the spectral distribution can be found by using Wien's displacement law:

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T is the absolute temperature

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So, the corresponding peak wavelength is

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A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from
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The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

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The acceleration of the car is calculated as;

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Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

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2 years ago
A gas‑forming reaction produces 1.90 m 3 1.90 m3 of gas against a constant pressure of 179.0 kPa. 179.0 kPa. Calculate the work
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Answer:

The work done is 3.4 × 10⁵ J.

Explanation:

Given:

Pressure of the gas produced (P) = 179 kPa

Volume of the gas produced (ΔV) = 1.90 m³

We need to find the work done in joules. For that, we don't need any conversion as the units are already in SI units which will give the result in Joules only.

Now, let us verify our results by using conversion factors and without using them.

Using conversion factors:

1 m³ = 1000 L

So, 1.90 m³ = 1.90 m³ × 1000 \frac{L}{m^3} = 1900 L

Also, 1 atm = 101.325 kPa

So, 179 kPa = 179 kPa × \frac{1\ atm}{101.325\ kPa} = 1.767 atm

Now, work done in a constant pressure process is given as:

Work = Pressure × Volume change

Work = P × ΔV

Work = 1.767 atm × 1900 L

Work = 3.36 × 10³ atm-L

Now, again using the energy conversion for work.

1 atm-L = 101.325 J

So, 3.36 × 10³ atm-L = 3.36 × 10³ atm-L × \frac{101.325\ J}{1\ atm-L} = 3.4 × 10⁵ J

Therefore, the work done is 3.4 × 10⁵ J.

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Work = P × ΔV = 179 × 1000 × 1.90 = 3.4 × 10⁵ J.

Therefore, the work done is same by both ways.

Hence the work done is 3.4 × 10⁵ J.

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