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Helen [10]
3 years ago
9

Identical 4.0-uC (microCoulomb) charges are placed on the y axis at y = +/-4.0 m. (plus or minus 4.0 m) Point A is on the x axis

at x = +3.0 m. Determine the electric potential of point A (relative to zero at the origin).
Physics
1 answer:
const2013 [10]3 years ago
4 0

Answer:

Explanation:

Distance of charges placed on y axis from given point on x axis

= √ (3² + 4² )

= 5 m

Potential at point A due to given charge

= Q / 4πε₀ R

Potential due to each of charges

= 4 x 10⁻⁶ x 9 x 10⁹ / 5  [ 1/ 4πε₀ = 9 x 10⁹ ]

= 7.2 x 10³ V

Potential due to both the charges

= 2 x 7.2 x 10³

= 14.4 kV .

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A very good insulator has an R-value of 29. The material's heat transfer coefficient is
Fittoniya [83]

Answer:

b. 0.034

Explanation:

The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

U = \frac{1}{R}

where

R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

U=\frac{1}{29}=0.034

4 0
3 years ago
~~~~~NEED HELP ASAP~~~~~
Romashka-Z-Leto [24]

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

8 0
3 years ago
Adult men have an average height of 69.0 inches with standard deviation of 2.8 inches fins the night of a man with a z-score of
ANTONII [103]
What do you mean I’m confused
7 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 7.1 x 10-9 m2 and a thickness of 1.5 x 10-8 m. A
hoa [83]

Answer:

Q = +1.4 pC

Explanation:

  • The capacitance of any capacitor, by definition, is as follows:
  •  C = \frac{Q}{V} (1)
  • Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
  •  C = \frac{\epsilon_{r}*\epsilon_{0} * A}{d} (2)
  • Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
  • Replacing by the givens in (2) we can find the value of the capacitance C, as follows:

       C = \frac{\ 4.4*8.85e-12C2/N*m2*7.1e-9m2}{1.5e-8m} = 18.4e-12 F = 18.4 pF

  • Replacing the values of C and V in (1), we can solve for Q, as follows:

       Q = C*V = 18.4e-12 F* 75.9e-3 V = 1.4e-12 C = +1.4 pC

  • As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.
7 0
4 years ago
I have two questions
geniusboy [140]
Current = voltage / resistance = 120 / 24 = 5 Amperes

Voltage = current x resistance = 6 x 10 = 60 Volts
5 0
3 years ago
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