All categories

2 years ago

What do you mean by magnitude of an electric field due to a point charge. State its SI unit.

1 answer:

3 0

Electric field strength is the force acting on a 1 Coulomb of positive change at that point. It is measured in Newton per Coulomb.

You might be interested in

Most asteroids lie between the orbits of

Anika [276]

I believe the answer is B. that's where the asetroid belt is.

4 0

2 years ago

Read 2 more answers

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7

Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I= $\frac{P_{avg}}{A}$

I= $\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}$

I= $7.296\times 10^{-6} W/m^2$

Amplitude of electric field at receiver end

$E_{max}=\sqrt{2I\mu _0c}$

Amplitude of induced emf

= $E_{max}d$

= $\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75$

= $17.591\times 10^{-2}=0.1759 v$

7 0

3 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

A turtle ambles leisurely, as turtles tend to do, when it moves from a location with position vector 1,=1.91 m and 1,=−2.73 m in

Elena-2011 [213]

Answer:

Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s

Explanation:

The displacement in the x-direction is:

$d_x = 3.65-1.91=1.74 m$

While the displacement in the y-direction is:

$d_y = -4.79 -(-2.73)=-2.06 m$

The time taken is t = 304 s.

So the components of the average velocity are:

$v_x = \frac{d_x}{t}=\frac{1.74}{304}=0.0057 m/s$

$v_y = \frac{d_y}{t}=\frac{-2.06}{304}=-0.0068 m/s$

And the magnitude of the average velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(0.0057)^2+(-0.0068)^2}=0.0089 m/s$

8 0

3 years ago

The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.

Natali [406]

Answer:

1500 divided by 150(15m x 10m/s^2) = 10

8 0

2 years ago