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schepotkina [342]
3 years ago
14

A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106

N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P
Physics
1 answer:
LenKa [72]3 years ago
8 0

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

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the SI and CGS unit of force are newton and dyne respectively haow many dynes are equal to one newton​
Triss [41]

Answer:

For example, the CGS unit of force is the dyne, which is defined as 1 g⋅cm/s2, so the SI unit of force, the newton (1 kg⋅m/s2), is equal to 100000 dynes.

Explanation:

5 0
3 years ago
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p
Rufina [12.5K]

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁  = (A₂ / A₁)v₂

v₁  = (0.00159 m²/ 0.0028278  m²) v₂ = 0.562  v₂

substitute v₁  into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂²  = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

v₂ = √ 20.472 = 4.525 m/s

volume follow rate = 0.00159 m² ×  4.525 m/s = 0.0072 m³/s

8 0
3 years ago
Why are all galaxies moving away from us?
ipn [44]
Galaxy million of star and planet. gravitional wave field all the universe some planet explosive itself moving other places . Black holes Mass gravity field
7 0
3 years ago
Given a 10-V power supply, would a 20-ohm resistor and a 5-ohm resistor need to be arranged in parallel or in series to generate
Bogdan [553]

Answer:

The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

Explanation:

We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.

We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.

Using ohm's law we have

V = I*R

V= voltage

I= current

R= resistance

10 = 2.5*R

R = 10/2.5 = 4ohm

that means we need a total of 4ohm resistance from these two resistors.

since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

\frac{1}{R} = \frac{1}{R1} +\frac{1}{R2}\\                 = \frac{1}{20} +\frac{1}{5}\\ \\                 = \frac{1+4}{20} =\frac{1}{4}

R(net) =4ohm

Now the orientation of the resistors are in parallel so the current will be divided.

we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.

We know that the voltage in parallel remains same

In 20 ohm resistance

again using ohms law

V = i1*R1

10 = i1*20

i1 = 0.5A

in 5ohm resistor

V=i2*R2

10 = I2*5

i2 =2A

and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.

Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

6 0
2 years ago
A certain resistance thermometer read 14.5 ohms in pure melting ice and 18.5 ohms in steam at standard atmospheric pressure what
Vadim26 [7]

The resistance of the thermometer at room temperature is 15.04 ohms.

<h3 />

<h3>What is a resistance thermometer?</h3>

A resistance thermometer is a type of thermometer that measures temperature through a change in resistance.

To calculate the resistance of the thermometer at room temperature, we use the formula below.

Formula:

  • 100/27 = 2/(x-14.5)..............Eqquation 1

Where:

  • x = Resistance of the thermometer at room temperature

Make x the subject of the equation

  • x = [(27×2)/100]+14.5
  • x = (54/100)+14.5
  • x = 0.54+14.5
  • x = 15.04 ohms.

Hence, The resistance of the thermometer at room temperature is 15.04 ohms.

Learn more about thermometers here: brainly.com/question/1531442

3 0
2 years ago
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