A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106 N/C. If the test charge is replaced with another test charge of 23 mC, what happens to the external electric field at P

Answers

Answer 1

Answer:

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

$E_2 = (E_1q_1)/(q_2)$

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

$E_2 = (4*10^6*13)/(23) = 2.26 *10^6 \ N/C$

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

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An electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. the electric field has amplitude 7.20×10−3v/m.what is the intensity of the wave in a medium\?

Answers

The intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².

What is the intensity of the wave?

The intensity of a wave is the total power delivered per unit area. It can be given as,

$I=(P)/(A)$

It can also be given as,

$I=(E^2)/(2)\sqrt{(k\varepsilon_o)/(\mu_r\mu_o)}$

Here, ( $\mu_r$ ) is relative permeability, ( $\mu_0$ ) is physical constant, (k) is dielectric constant, (E) is the amplitude of electric field, and $\varepsilon_o$ is the permittivity of free space.

Here, the electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency.

As the electric field has amplitude 7.20×10−3v/m. Thus, put the values in the above formula to find the intensity as,

$I=((7.20*10^(-3))^2)/(2)\sqrt{(3.64*8.85*10^(-12))/(5.18*(4\pi*10^(-7)))}\nI=5.766*10^(-8)\rm W/m^2$

Hence, the intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is 5.766×10⁻⁸ W/m².

Learn more about the intensity of the wave here;

brainly.com/question/18109453

Ans: Intensity = I = $5.8 * 10^(-8) W/m^2$

Explanation:
First you need to find out the speed of Electromagnetic wave:

Since
v = $\sqrt{ (1)/(\mu\epsilon) }$

v = $\sqrt{ (1)/(\mu_r\mu_ok\epsilon_o) }$
$\mu_r = 5.18 \n\mu_o = 4 \pi * 10^(-7) \nk = 3.64 \n\epsilon_o = 8.85 * 10^(-12)$

Plug in the values:
v = $\sqrt{ (1)/((5.18)(4\pi*10^(-7))(3.64)(8.85*10^(-12))) }$
v = $6.9 * 10^7$ m/s

Now that we have "v", we can use the following formula to find the intensity of wave:

$I = (k\epsilon_o*v*E_(max)^2)/(2)$

$I = ((3.64)(8.85*10^(-12))*(6.9*10^7)*(7.20*10^(-3))^2)/(2)$

Intensity = I = $5.8 * 10^(-8) W/m^2$

If you rub a balloon on your hair, youcan hang the balloon on the wall.
Why does the balloon stick to the
wall?

Answers

Answer:

The balloon is electrostatically charged

Explanation:

After rubbing it on the hair, the balloon is electrically charged, and as such, when approaching the wall it draws opposite charges from the wall creating locally on the wall's surface an accumulation of the charges opposite to the balloon . and repelling deeper into the wall those charges of the same sign.

A car traveling at 45 km/h starts to brake, and comes to a stop over a distance of 18 m. Calculate the accelerationof the braking car.

Answers

Answer:

Acceleration, $a=8.68\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 45 km/h = 12.5 m/s

Final speed, v = 0 (as they comes to rest)

Distance, d = 18 m

We need to find the acceleration of the breaking car. Using third equation of motion as follows :

$v^2-u^2=2ad\n\n\text{Where a is acceleration of the car}\n\na=(v^2-u^2)/(d)\n\na=((12.5)^2)/(18)\n\na=8.68\ m/s^2$

So, the acceleration of the braking car is $8.68\ m/s^2$ .

Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?

Answers

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = $3* 10^8\ m/s$

b = Wien's displacement constant = $2.897* 10^(-3)\ mK$

T = Temperature

From Wien's law we have

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(2900)\n\Rightarrow \lambda_m=9.98966* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(9.98966* 10^(-7))\n\Rightarrow \nu=3.00311* 10^(14)\ Hz$

For Halogen

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(3400)\n\Rightarrow \lambda_m=8.52059* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(8.52059* 10^(-7))\n\Rightarrow \nu=3.52088* 10^(14)\ Hz$

The maximum frequency is produced by Halogen bulbs which is closest to the value of $5.5* 10^(14)\ Hz$

Ratio

$(3.00311* 10^(14))/(3.52088* 10^(14))=0.85294$

The ratio of Incandescent to halogen peak frequency is 0.85294

A mysterious object with a surface area of 0.015 m2, volume of 0.000125 m3, density of 100 kg/m3, specific heat of 100 J/(kgK), thermal conductivity of 2 W/(mK), with an unknown initial temperature was placed in a fluid with a density of 50 kg/m3, specific heat of 70 J/(kgK), thermal conductivity of 0.1 W/(mK), at a temperature of 400K. The heat transfer coefficient is given to be 10 W/(m2K). After 10 seconds, the temperature of the object is measured to be 380K. Determine the object's initial temperature.

Answers

Answer:

The object's initial temperature is 333.6 K

Explanation:

We first assume that the liquid can only transfer heat to the object through convective heat transfer method.

Let T₀ = the initial temperature of the object

T = temperature of the object at anytime.

The rate of heat transfer from the liquid to the object is given as

Q = -hA (T∞ - T)

T∞ = temperature of the fluid = 400 K

A = Surface area of the object in contact with the liquid = 0.015 m²

h = Convective heat transfer coefficient is given to be = 10 W/(m²K)

The rate of heat gained by the object is given by

mC (d/dt)(T∞ - T)

m = mass of the object = ρV

ρ = density of the object = 100 kg/m³

V = volume of the object = 0.000125 m³

m = ρV = 100 × 0.000125 = 0.0125 kg

C = specific heat capacity of the object = 100 J/(kgK)

The rate of heat loss by the liquid = rate of heat gain by the object

-hA (T∞ - T) = mC (d/dt)(T∞ - T)

(d/dt)(T∞ - T) = - (dT/dt) ( Since T∞ is a constant)

- mC (dT/dt) = -hA (T∞ - T)

(dT/dt) = (hA/mC) (T∞ - T)

Let s = (hA/mC)

(dT/dt) = -s (T - T∞)

dT/(T - T∞) = -sdt

Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -st

(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ

(T - T∞) = (T₀ - T∞)e⁻ˢᵗ

s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12

T = 380 K at t = 10 s

T₀ = ?

T∞ = 400 K

st = 0.12 × 10 = 1.2

(380 - 400) = (T₀ - 400) e⁻¹•²

(-20/0.3012) = (T₀ - 400)

(T₀ - 400) = - 66.4

T₀ = 400 - 66.4 = 333.6 K

Hope this Helps!!!

When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery