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rewona
3 months ago
15

A uniformly charged rod (length = 2.0 m, charge per unit length = 5.0 nc/m) is bent to form one quadrant of a circle. what is th

e magnitude of the electric field at the center of the circle?
Physics
1 answer:
inna [3.1K]3 months ago
4 0

The electric field at the center of a circular arc can be determined using the formula

E = \frac{2k\lambda sin\frac{\theta}{2}}{R}

and we know that

\lambda = 5nC/m

k = 9 \times 10^9 Nm^2/C^2

\frac{\pi}{2}R = L = 2m

R = \frac{4}{\pi}

furthermore, we are aware that

\theta = \frac{\pi}{2}

now applying the formula above

E = \frac{2(9\times 10^{-9})(5\times 10^{-9})sin45}{4/\pi}

E = 50 N/C

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Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The
Keith_Richards [3271]

Answer:

11.56066 m/s

Explanation:

m = Mass of individual

v = Velocity of individual = 13.4 m/s

g = Gravitational acceleration = 9.81 m/s²

v' = Velocity of the individual after dropping

At the surface, kinetic and potential energy will equalize

\dfrac{1}{2}mv^2=mgh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{13.4^2}{2\times 9.81}\\\Rightarrow h=9.15188\ m

The cliff's height is 9.15188 m

Define fall height as h' = 2.34 m

\dfrac{1}{2}mv'^2+mgh'=mgh\\\Rightarrow v'=\sqrt{2g(h-h')}\\\Rightarrow v'=\sqrt{2\times 9.81(9.15188-2.34)}\\\Rightarrow v'=11.56066\ m/s

The person's speed is 11.56066 m/s

3 0
3 months ago
Add a third force that will cause the object to remain at rest. Label the new force F⃗ 3. Draw the vector starting at the black
kicyunya [3294]
The new force F3 is added in the same direction as F2. To analyze the forces acting on an object in this scenario, we observe that they operate along the vertical axis, with F1 acting upward and F2 downward. To determine the necessary vector F3 to counteract the net force, it's important to calculate the length difference between F1 and F2. The direction of F3 will match that of the smaller force. If F2 is less than F1, F3 will align with F2.
4 0
2 months ago
How many electrons must be removed from a neutral, isolated conducting sphere to give it a positive charge of 8.0 x 10 8 C? [Q=n
inna [3103]

The new charge of the ball will amount to 8x10^8C after removing 5x10^27 electrons.

Explanation:

Initially, if the sphere is electrically neutral, its charge stands at 0C.

When an electron with a charge of (-1.6*10^-19 C) is taken away, we effectively add a positive charge, leading to:

1.6*10^-19 C as the sphere's new charge.

For a total of N electrons removed, the sphere's overall charge now becomes:

N*1.6*10^-19 C.

To calculate N when:

N*1.6*10^-19 C = 8.0x 10^8 C.

We find that N is: (8.0/1.6)x10^(8 + 19) = 5x10^27 electrons.

7 0
4 months ago
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