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zaharov [31]
3 years ago
5

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

of the electric field at the center of the circle?

Physics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

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a child hits a ball with a force of 350 N. (a) If the ball and bat are in contact for 0.12 is, what impulse does the ball receiv
Lina20 [59]

Explanation:

Given that,

Force with which a child hits a ball is 350 N

Time of contact is 0.12 s

We need to find the impulse received by the ball. The impulse delivered is given by :

J=F\times t\\\\J=350\times 0.12\\\\J=42\ N-m

So, the impulse is 42 N-m..

We know that he change in momentum is also equal to the impulse delivered.

So, impulse = 42 N-m and change in momentum =42 N-m.

7 0
3 years ago
Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge
asambeis [7]

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

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

q =\dfrac{Er^2}{k}

inserting all the values

q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}

      q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0
3 years ago
Acceleration problem <br> Show work plz
Dennis_Churaev [7]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)  

vo² = 630               simplifying

vo = 25 m/s            result

3 0
3 years ago
Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each
Hatshy [7]

Answer:

forcing in act

Explanation:

7 0
3 years ago
Help please thank you
trasher [3.6K]
I believe the answer would be C because potential energy is affected by height and mass. The truck in photo C is the highest and has a lot of mass.
4 0
3 years ago
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