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serg [7]
3 years ago
5

A +26.3 uC charge qy is repelled by a force

Physics
1 answer:
Musya8 [376]3 years ago
5 0

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

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AlekseyPX
It is a chemical compound
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3 years ago
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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2
Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0
4 years ago
Physical science
attashe74 [19]

Answer:

0.56 km/s

Explanation:

We will define a single system of units for measurement, for this case meters per second [m/s]. That is, we must convert the rest of units such as centimeters per second and kilometers per second to meters per second.

560[\frac{cm}{s}]*(\frac{1m}{100cm} )=5.6[m/s]\\0.56[\frac{km}{s}]*(\frac{1000m}{1km} )=560[m/s]

Therefore the speed of 0.56 [km/s] is the greatest of all

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4 years ago
In an inelastic collision involving an isolated system, the final total momentum isA) more than the initial momentum.B) exactly
damaskus [11]

I think that the answer is C

5 0
3 years ago
A gas mixture contains 320mg methane, 175 mg argon, 225 mg nitrogen (N2). The partial pressure of argon at 300K is 12.52 kPa. Wh
svet-max [94.6K]

<u>Answer:</u> The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 320 mg = 0.3 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol

  • <u>For argon:</u>

Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol

  • <u>For nitrogen:</u>

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol

To calculate the volume of the mixture, we use the equation:

PV = nRT         ......(2)

We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

5 0
3 years ago
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