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2 years ago

13

What is the formula to find the second charge in Coulomb's Law?

2 answers:

kicyunya [14]2 years ago

5 0

Answer:

1.F is the electrostatic force between charges (in Newtons),

2.q₁ is the magnitude of the first charge (in Coulombs),

3.q₂ is the magnitude of the second charge (in Coulombs),

4.r is the shortest distance between the charges (in m),

5.ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C² .

crimeas [40]2 years ago

4 0

Explanation:

F is the electrostatic force between charges (in Newtons), 2. q, is the magnitude of the first charge (in Coulombs), 3. q2 is the magnitude of the second charge (in Coulombs), 4. r is the shortest distance between the charges (in m), 5. ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N-m²/C².

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saveliy_v [14]

Answer:

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Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

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