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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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Describe how a change in resistance would affect the current in a circuit.
Tresset [83]

Explanation:

The relation between potential difference, current and resistance flowing in a circuit is given by using Ohm's law. It can be given by :

V = IR

I=\dfrac{V}{R}

Resistance opposes the flow of electric current in the circuit. It means that, if resistance is more, less current will flow through the circuit.

3 0
3 years ago
An athlete whose mass is 97.0 kg kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with cons
SIZIF [17.4K]

Answer:

470 N.

Explanation:

Using equations of motion:

S = vi*t + 1/2*(a*(t^2))

Given:

S = 0.65 m

t = 1.5 s

vi = 0 m/s

0.65 = 1/2 * (a * (1.5)^2)

a = 1.3/2.25

= 0.578 m/s^2

Force = mass * acceleration due to gravity

= 92 * 0.578

= 53.16 N

Total force = 420 + 53.16

= 473.16 N

= 470 N.

6 0
3 years ago
Approximate the field by treating the disk as a +6.1 C point charge at a distance of 21 cm Answer in units of N/C.
DaniilM [7]

Answer:

Electric field, E=1.24\times 10^{12}\ N/C

Explanation:

It is given that,

Charge, Q = +6.1 C

Distance, r = 21 cm = 0.21 m

We need to find the electric field. It is given by :

E=k\dfrac{Q}{r^2}

E=9\times 10^9\times \dfrac{6.1}{(0.21)^2}

E=1.24\times 10^{12}\ N/C

So, the electric field at this distance is 1.24\times 10^{12}\ N/C. Hence, this is the required solution.

7 0
3 years ago
32. (FR) A mass moving at 25 m/s slides along a rough horizontal surface. The coefficient of friction is 0.30. (A) Use force and
scoray [572]

Answer:

A)s = 104.16 m

b)s= 104.16 m

Explanation:

Given that

u = 25 m/s

μ = 0.3

The friction force will act opposite to the direction of motion.

Fr= μ m g

Fr= -  m a

a=acceleration

μ m g = - m a

a= - μ g

a= - 0.3 x 10 m/s²          ( take g= 10 m/s²)

a= - 3  m/s²

The final speed of the mass is zero ,v= 0

We know that

v² = u² +2 a s

s=distance

0² = 25² - 2 x 3 x s

625 = 6 s

s = 104.16 m

By using energy conservation

Work done by all the forces =Change in the kinetic energy

- Fr.s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Negative sign because force act opposite to the displacement.

- \mu\ m\ g \ s=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

-\mu\ g \ s=\dfrac{1}{2}v^2-\dfrac{1}{2}u^2

-0.3\times 10\times \ s=\dfrac{1}{2}\times 0^2-\dfrac{1}{2}\times 25^2

- 3 x 2 x s = - 625

s= 104.16 m

4 0
3 years ago
If your speedometer has an uncertainty of 2.5 km/h at a speed of 92 km/h, what is the percent uncertainty
Elis [28]

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = \frac{u}{v} * 100%

p = \frac{2.5}{92} * 100%

p = 2.7%

Therefore, the percent uncertainty is 2.7%

6 0
3 years ago
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