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kakasveta [241]
3 years ago
8

A uniformly charged sphere has a total charge of 300uc and a radius of 8cm . Find the electric field density at a point on the s

urface of the sphere​
Physics
2 answers:
algol133 years ago
7 0
The answer above is right
liubo4ka [24]3 years ago
3 0

Answer:

1.05×108NC−1,1.17×108NC−1, zero

Explanation:

A uniformly charged sphere has a total charge of 300muC and radius of 8 cm. find the electric field (i) at a point 16 cm from the centre of the sphere (ii) at a point on the surface of the sphere (iii) at a point inside the sphere.

CARRY ON LEARNING

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Kobotan [32]

Answer:

<h2><em>3.067MHz</em></h2>

Explanation:

The formula for calculating the voltage across an inductor is expressed as

V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)

Given parameters

current amplitude I = 1.50mA = 1.5*10⁻³A

inductance L = 0.450mH = 0.450*10⁻³H

Voltage across the inductor V_l = 13.0V

Required

frequency f

Substituting the given parametres into the formula, we have;

V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz

<em>Hence, the frequency required is 3.067MHz</em>

4 0
3 years ago
Unit and dimension of luminous intensity​
DanielleElmas [232]

Answer:

unit - candela (cd)

Dimension - J

8 0
3 years ago
PLZ HELP ME GUYS!!!! Using appropriate terminology, contrast transverse and longitudinal waves. Hint: How are they different? Yo
kobusy [5.1K]
<h3>Longitudinal Waves:</h3>
  1. Longitudinal waves are waves in which particles moves in the direction parallel to the direction of wave.
  2. Longitudinal wave are one directional waves.
  3. Sound waves and tension force applied on spring are examples of longitudinal waves.
<h3>Transverse Waves:</h3>
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  3. Ripples formed on water and all electromagnetic waves are examples of transverse waves.
6 0
4 years ago
a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?​
jolli1 [7]

Answer:

Elevation =31.85[m]

Explanation:

We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.

We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.

The point where the ball is thrown shall be taken as a reference point of potential energy.

E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J]

m = mass of the ball = 300 [gr] = 0.3 [kg]

v = initial velocity = 25 [m/s]

E_{k}=\frac{1}{2}  * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J]

93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m]

5 0
4 years ago
4. Which point on this position vs. time graph has the fastest speed?
galben [10]

Answer:

Since v = (x(2) - x(1)) / t

point 2 obviously has the greatest displacement in a given time

Also, point 2 is the steepest line on this graph.

3 0
3 years ago
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