Ask question

Ask question

All categories

3 years ago

13

An electron beam moves toward a cathode ray tube screen, which is 50 cm away from the negative electrode. The electrons are acce

lerated by a potential difference of 18 kV. Estimate the maximum displacement of the electron beam caused by Earth's magnetic field. The average magnetic field at the surface of Earth is roughly BE = 45×10−6T. Δz = ?

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

Maximum displacement of the Electron beam= 0.0124m

Explanation:

Detailed solution of the question is given in the attached files.

You might be interested in

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has

Anastasy [175]

-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is

   (3.95 x 10⁵m) + (4.2 x 10⁶m)

   = (3.95 x 10⁵m) + (42 x 10⁵m)

   = 45.95 x 10⁵ m

The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

   (91.9 π x 10⁵ m) / 2 hr

= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)

=    0.04 x 10⁵ m/sec

      =    4 x 10³    m/sec

   (4 kilometers per second)

6 0

3 years ago

For the circuit in the previous part, what happens to the maximum current if the frequency is doubled and the inductance is halv

tamaranim1 [39]

Answer:

Following are the responses to these question:

Explanation:

Since the $max^{m}$ is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance $(X_L)$ for capacities

$\to X_{C}=\frac{1}{W L}$

for

$\to X_{L}=wL$

When $w \longrightarrow 2w$

$L\longrightarrow \frac{L}{2}$

then

$\to X_{L}=2 w \times \frac{L}{2}=wL$

therefore, $X_{L}$ remains at the same so, the maximum current remains the in same ckt.

4 0

3 years ago

A particle is traveling in the positive direction along an x axis, at a constant 5 m/s. Which of the following

kvasek [131]

The particle will accelerate 5m/s every second until it reaches a maximum of whatever your graph/diagram goes to, I'm in physical science and this is somewhat similar to what I am doing now but I'm not sure if that was what your looking for.

8 0

3 years ago

A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m

kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm$

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm$

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0

3 years ago