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2 years ago

Conventionally, the field strength around a charged object is the direction of the force acting on a .

Physics

2 answers:

damaskus [11]2 years ago

8 0

Answer:

Field Strength:

Any charged body has the capacity to effect any test charge that comes inside its field or region. It can be also defined as,

"The total amount or magnitude of force,F or intensity felt by any unit test charge when it enters an electromagnetic or electric field."

For an electric field its unit will be, volt per meter or simply V/m.

Explanation:

A unit test charge inside an electric field:

When a unit test charge enters a given parameters or area set by the charged particle then it will surely experience a force,F equal to the magnitude of that charge body and it will be different as it continues to move closer or far inside the field.

As, we have, "E=F/q",(where "F" is the field strength and "q" is the unit test charge placed inside the field).

ExtremeBDS [4]2 years ago

7 0

Unit positive test charge

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A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

$\theta _2 = 13^o$

$\theta _1 =32.94^o$

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

2 years ago

g A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.4 A when an additional 2.2 Ω resistor

Mrrafil [7]

Answer:

R1 = 5.13 Ω

Explanation:

From Ohm's law,

V = IR............... Equation 1

Where V = Voltage, I = current, R = resistance.

From the question,

I = 2 A, R = R1

Substitute into equation 1

V = 2R1................ Equation 2

When a resistance of 2.2Ω is added in series with R1,

assuming the voltage source remain constant

R = 2.2+R1, and I = 1.4 A

V = 1.4(2.2+R1)................. Equation 3

Substitute the value of V into equation 3

2R1 = 1.4(2.2+R1)

2R1 = 3.08+1.4R1

2R1-1.4R1 = 3.08

0.6R1 = 3.08

R1 = 3.08/0.6

R1 = 5.13 Ω

6 0

2 years ago

"Calculate the speed of this sound wave by recording how much time it takes to travel 5 meters. Use the (old school) velocity eq

lianna [129]

1) sound velocity reported by you : 292.39 m /s

2) time to travel 1620m at that velocity: t = d / v = 1620 m / 292.39 m/s = 5.54 s, since the moment the sound wave started.

3) You might wanted to tell the time since you watched the lightning.

Then you can calculate the time since the lighting was generated,1620 m away from you, until you saw it, using the speed of light:

speed of light = 3*10^8 m/s => t = 1620 m / (3*10^8m/s) =0.0000054 s

Then, this time is completely neglectible, and yet the answer is 5.54 s, as calculated in the step 2.

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2 years ago

What is the kinetic energy of a 7.56 kg bowling ball that is rolling at a speed of 2 m/s ?

postnew [5]

Answer:

K.E=?

Mass=7.56 kg

Velocity=2m/s

Solution::

K.E = 1/2mv^2

K.E=0.5*7.56*(2)^2

K.E= 0.5*7.56*4

K.E=15.12 joules.

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2 years ago

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1 year ago