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Ksivusya [100]
3 years ago
7

An unknown additional charge q3 is now placed at point B, located at coordinates (0 m, 15.0 m ). Find the magnitude and sign of

q3 needed to make the total electric field at point A equal to zero.

Physics
1 answer:
Finger [1]3 years ago
5 0

The first part of the question is:

Two point charges are placed on the x axis. (Attached file)The first charge, q1 = 8.00 nC , is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC , is placed a distance 9.00 m from the origin along the negative x axis.

Answer:

q3 = +0.3nc

Explanation:

Due to the vector symbols in the solution, I've decided to attach the explanation to this answer.

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A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat
Triss [41]

Answer:

2.1406 ×10^6 m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

qV=\frac{1}{2}mv^2

here V is the potential difference  

we know that mass of proton = 1.67×10^{-27} kg

we have given speed =50000m/sec

so potential difference V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045

now mass of electron =9.11×10^{-31}

so for electron

\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec

so the velocity of electron will be 2.1406×10^6 m/sec

4 0
3 years ago
A 60-watt light bulb has a voltage of 120 bolts applied across it and a current of 0.5 amperes flows through the bulb. What is t
Andre45 [30]

Answer:

240 ohms

Explanation:

From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then

R=120/0.5=240 Ohms

Alternatively, resistance is equal to voltage squared divided by watts hence \frac {120^{2}}{60}=240

7 0
3 years ago
At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

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

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0
3 years ago
Which term is used to describe the variety of inheritable traits in a species?
choli [55]
Genetic i think is it right?
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Which branch of physics deals with the study of force, energy, and motion?
dexar [7]
The correct answer is Mechanics
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