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3 years ago

Two point charges of magnitude 5.0 nC and -3.0 nC are separated by

Physics

1 answer:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

V = 411.43 V

Explanation:

The two forces as a result of each of the 2 charges are;

F1 = kq1•q/r

F2 = kq2.q/r

Where r = r/2 since we are dealing with potential difference at a point midway between the charges.

q1 = 5 nC = 5 × 10^(-9) C

q2 = 3 nC = 3 × 10^(-9) C

k = 9 × 10^(9) N.m²/C²

r = 35 cm = 0.35m

r/2 = 0.35/2

Thus;

F1 = (9 × 10^(9) × 5 × 10^(-9) × q)/(0.35/2)²

F1 = 1469.39q

F2 = (9 × 10^(9) × 3 × 10^(-9) × q)/(0.35/2)²

F2 = 881.63q

Net force acting midway is;

F_net = F1 + F2

F_net = 1469.39q + 881.63q

F_net = 2351.02q

Now, we know that formula for electric potential is;

V = kq/r

Thus ;

V = Fr/q derived from the earlier equation for force we used.

Where F is F_net.

V = 2351.02q × r/q

V = 2351.02r

Recall that we are dealing with midpoint and r = r/2

Thus;

V = 2351.02 × 0.35/2

V = 411.43 V

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Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago

Suppose Mary would like to design an emergency overheating alarm using one of these materials. In her design, at room temperatur

Reptile [31]

Answer:

A. 92.88 °C

B. 401.535 °C

Explanation:

So, the overheating system is going to be based on the principle of thermal linear expansion. The behavior of this principle is ruled by the equation:

∆L= L_i * ∆T * α

Where α is a coefficient of linear expansion. For a cylinder made from polycarbonate α = 70,2*10^(-6) °C^(-1) and for a cylinder made from cast iron α = 12*10^(-6) °C^(-1). If we isolate the term of the temperature’s difference, we have:

∆L/(L_i * α) = ∆T → T_f = T_i + ∆L/(L_i * α)

Replacing the values, for the case of the Polycarbonate we have:

T_f = T_i + ∆L/(L_i * α) = 23°C+0,0273cm/(6,01cm *70,2 * 10^(-6)°C^(-1) ) = 92,88 °C

Replacing the values, for the case of the Cast Iron we have:

T_f = T_i +∆L/(L_i * α) = 23°C + 0,0273cm/(6,01cm * 12 * 10^(-6) °C^(-1) ) = 401,535 °C

As we see, is way better to use the polycarbonate in this application.

Have a nice day. Let me know if I can help you with anything else. :D

4 0

3 years ago

Is gravity a property of mass ?

Alex787 [66]

Answer:

no it is not a property of mass

5 0

3 years ago

Read 2 more answers

Which of the following is NOT a requirement for a planet? A. must orbit a star, but is not a star or satellite of another planet

Ket [755]

Answer:

D. must have a moon

Explanation:

Scientists say that a planet must orbit a star, must be round, and must be big enough so its gravity clears its orbit of other objects. Therefore, the statement that is not a requirement for a planet is

D. must have a moon

For example, Mercury and Venus are planets with no moons.

6 0

1 year ago

PHYSICS, HELP PLZ??!! 100PTS

romanna [79]

Hi there!

We can begin by calculating the time the ball takes to reach the highest point of its trajectory, which can be found using the following:

$t_{max} = \frac{vsin\theta}{g}$