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Irina-Kira [14]
2 years ago
15

two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

ield at the midpoint between the two charges?
Physics
1 answer:
Scorpion4ik [409]2 years ago
6 0

Answer:

Electric field due to two charges is given as

E = 1.44 \times 10^7 N/C

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

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

now we have

q = 2\times 10^{-6} C

r = 5 cm = 0.05 m

now we have

E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}

E = 1.44 \times 10^7 N/C

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Find the current passing through each of the 3 resistors connected parallel to each other as shown in the figure (i1, i2, I3). S
Thepotemich [5.8K]

Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Explanation:

From the image, we see that the resistors are connected in parallel. This means that the voltage passing through them is the same.

Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

Thus;

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

3 0
3 years ago
A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe
mezya [45]

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

6 0
3 years ago
One of the BEST measures of success for businesses and their entrepreneurs is
Sidana [21]
Are there choices or no??
6 0
3 years ago
Read 2 more answers
A car battery seems to be malfunctioning (not providing the proper voltage and current to start your car). While the car is off
Stella [2.4K]

No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.

4 0
2 years ago
A "590-W" electric heater is designed to operate from 120-V lines.
lukranit [14]

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

d)

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

b)

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

<u>I = 4.92 A</u>

<u></u>

a)

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

<u>R = 24.4 Ω</u>

<u></u>

c)

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

<u>P = 495.9 W</u>

<u></u>

d)

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

<u>c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.</u>

7 0
2 years ago
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