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garri49 [273]
3 years ago
9

A point charge 20.0 nC is located at the center of a cube whose sides are of length 10.0 cm. If there is no other charges in thi

s system, what is the electric flux through one face of the cube?
Physics
1 answer:
stealth61 [152]3 years ago
3 0

Answer:

ϕ_Net = 2258.87 Nm²/C

Explanation:

We are given;

Charge; q = 20 nC = 20 × 10^(-9) C

Length of each side; L = 10 cm = 0.1 m

The net electric flux through one face of the cube is given by the formula;

ϕ_Net = q/ε_o

Where ε_o is a constant known as vacuum of Permittivity and has a value of 8.854 × 10^(−12) C²/N.m²

Thus;

ϕ_Net = (20 × 10^(-9))/(8.854 × 10^(−12))

ϕ_Net = 2258.87 Nm²/C

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A beverage manufacturer wants to increase the solubility of carbon dioxide (CO2) in its carbonated drinks.
Brut [27]

D. Decreasing its temperature

Explanation:

Decreasing the temperature of the carbon dioxide gas to be dissolved in the carbonated drink will most likely increase the solubility of the gas in the drink.

Temperature has considerable effects on the solubility of gases in liquids.

  • Dissolution involves the surrounding of ions by water molecules, in this case, the carbon dioxide gas is to be surrounded by the liquid beverage medium.
  • Increasing pressure increases the rate at which gases are soluble. At high pressure, the gases are brought more in contact with the liquid medium.
  • Decreasing temperature aids gas solubility.
  • If the temperature of gases are increased,  they will not want to stay in solution as they gain a high amount of kinetic energy.
  • Therefore, it will increase their randomness and the urge to leave the solution.
  • Decrease in temperature and increase in pressure makes gas solubility to be fast.  

Learn more:

Rate of chemical reactions brainly.com/question/6281756

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6 0
3 years ago
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7
My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

4 0
3 years ago
Why do noble gasses rarely react with other elements?
ollegr [7]
Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.
3 0
2 years ago
A wave has a frequency of 60 Hz and a wavelength of 1.7 meters. What is the speed of this wave? equation, substitution
azamat

Answer:

v = 102 m/s

Explanation:

Given that,

The frequency of a wave, f = 60 Hz

Wavelength = 1.7 m

We need to find the speed of this wave. The formula for the speed of a wave is given by

v=f\lambda

Substitute all the values,

v=60\times 1.7\\\\v=102\ m/s

So, the speed of the wave is equal to 102 m/s.

4 0
3 years ago
The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.
Alexus [3.1K]

Answer:

(a) θ= 43.89°

(b) v_{1} = 1.88\sqrt{3} \frac{m}{s}

    v_{2} = 6.79 \frac{m}{s}

Explanation:

Ball 1:

u_{1} = 7.52\frac{m}{s}

Ball 2:

u_{2} = 0\frac{m}{s}

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

\frac{1}{2} m_{1}u_{1}^{2}   = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}

and

m_{1}u_{1} =   m_{1}v_{1} + m_{2}v_{2}

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

u_{1}^{2}   = v_{1}^{2} + v_{2}^{2}

and

u_{1} =   v_{1} + v_{2}

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

56.55 = v_{1} ^{2} + v_{2} ^{2}

and

u_{1} = v_{1}cos(30) +  v_{2}cos(\theta)

Solving we get:

(\frac{2u_{1}-\sqrt{3}v_{1}  }{2cos(\theta)})^{2}  = 56.55-v_{1} ^{2}

From conservation in y-direction, we get:

0 = v_{1}sin(30) -  v_{2}sin(\theta)

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.

8 0
3 years ago
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