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fomenos
3 years ago
8

Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric fiel

d has a magnitude of 50 N/C .
Physics
1 answer:
Delicious77 [7]3 years ago
4 0

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

E=2\frac{kQ}{r^2}

Rearranging for Q

Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}

Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC

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Which property describes the temperature at which a liquid changes to a gas?
Rufina [12.5K]

Answer:

Boiling Point

Explanation:

When a liquid changes to a gas is called the boiling point.

5 0
3 years ago
An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole
Helga [31]

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

<h2>ε = 2 V/cm</h2><h2></h2>

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

8 0
3 years ago
When looking at a graph what are the first 3 things you should do?
Valentin [98]

Answer: Look where the points are.

Explanation:

4 0
3 years ago
Read 2 more answers
Transformer is used in rectifier step up and step down voltage?<br>​
tensa zangetsu [6.8K]

Answer:

While the use of the type of transformer in a rectifier depends on the voltage requirement or to meet desired operating conditions, a step-down transformer is used mainly to reduce the voltage. It is used to bring the high AC voltage level to a reasonable value or the desired output voltage.

Explanation:

Hope it helps

Correct me if Im wrong

3 0
2 years ago
A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-
SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency \omega_0 at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

Where

L = Inductance

C = Capacitance

Our values are given as

C = 15*10^{-6}\mu F

L = 0.280*10^{-3}mH

Replacing we have,

f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})

f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})

f= 2455.81Hz

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0
3 years ago
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