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Katyanochek1 [597]
3 years ago
12

A 2.4 nC charge is at the origin and a -4.8 nC charge is at x = 1.3 cm . At what x-coordinate could you place a proton so that i

t would experience no net force?
Physics
1 answer:
MaRussiya [10]3 years ago
3 0
The proton (positive charge) shall be closer to the charge with the lower magnitude  which is at the orgin.

The proton also shall be out of the interval between the two charges so that the pull of one charge cancels with the push of the other.

The region at which those conditions happen is to the left of the origin.

In that case the forces over the proton shall be:

k* (2.4 nC) * p / (x^2) - k*(4.8 nC) * p /( 1.3 + x)^2 = 0


where p is the charge of the proton.

You can simplify k and p:

2.4 / x^2 - 4.8 / (1.3 + x)^2 = 0

You can also simplify by 2.4

1/ x^2 - 2 / (1.3 + x)^2 = 0

(1.3+x)^2 - 2x^2 = 0

1.69 + 2.6x + x^2 - 2x^2 = 0

1.69 + 2.6x - x^2 = 0

x^2 -2.6x - 1.69 =0

Solve using the quadratic formula: x = 3.14 (use only the positive value)

That is the proton shall be place 3.14 units to the left of the origin (positive charge) 

  


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A car's engine generates 100,000W of power as it exerts a force of 10,000N . How long does it take the car to travel 100m ?
kompoz [17]

Answer: 10 s

Explanation:

Given

Engine generates 100,000\ W power

Force exerts F=10,000\ N

Distance traveled d=100\ m

work done is given by

W=F\cdot d\\W=10,000\times 100\\W=10^5\ J

Also, Energy is given as

\Rightarrow E=P\cdot t

Insert the values

\Rightarrow 10^6=100,000\times t\\\\\Rightarrow t=\dfrac{10^6}{10^5}\\\\\Rightarrow t=10\ s

5 0
2 years ago
A proton is released in a uniform electric field, and it experiences an electric force of 2.36×10−14 N toward the south. (a)What
qwelly [4]

Answer:

a) E = 1.47 × 10^5 N/C

b) south

Explanation:

The magnitude of an electric field can be defined mathematically as;

E = F/q ........1

Where,

E = magnitude of the electric field

F = electric force

q = charge on the proton

Given;

F = 2.36 × 10^-14 N

Note that charge on a proton is known as Qp = 1.602 × 10^-19 C

q = 1.602 × 10^-19 C

Substituting into equation 1, we have;

E = 2.36 × 10^-14 N/1.602 × 10^-19 C

E = 1.47 × 10^5 N/C

b) The direction of the electric field;

From equation 1

E = F/q ........1

since both electric field and electric force are vector quantity and q is a positive charge (constant), then both the electric field and electric force would be parallel to each other. Therefore the electric field is directed to the south also.

(When a vector is multiplied by a positive constant the direction remains the same)

7 0
3 years ago
What decibel reading corresponds to a pressure amplitude of 0.2 W/m^2?
horsena [70]

I  = pressure amplitude given = 0.2 W/m²

dB = decibel reading

decibel reading from the pressure amplitude is given as

dB = 10 log₁₀ (I/10⁻¹²)

inserting the values in the above equation

dB = 10 log₁₀ (0.2/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹.10¹²)

dB = 10 log₁₀ (2 x 10¹²⁻¹)

dB = 10 log₁₀ (2 x 10¹¹)

dB = 113.01 db

hence the decibel reading comes out to be 113.01 db


4 0
3 years ago
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