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3 years ago

Consider the following kinds of electromagnetic waves:

Physics

1 answer:

marysya [2.9K]3 years ago

7 0

Answer:

According to wavelengths in descending order: C, B, D, E and A

According to frequencies in descending order: A, E, D, B and C

According to speeds in vacuum: A =B =C =D =E

Explanation:

in the EM spectrum, radio waves have the longest wavelength while gamma rays have the shortest wavelength.

All EM waves travel with the same speed of $3*10^8m/s$ in a vacuum.

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An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 7 m/s over

stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

= 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .

xf =xi+vi*t+1/2*a*t

t=√2(xf-xi)/a

t=3.765×10^-10 s

now we can use the power P Eq.

P=W/Δt => ΔK/Δt

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0

2 years ago

an Olympic runner leaps over a hurdle.if the runner initial vertical speed is 2.2m/s how much will the runners center of mass be

Snowcat [4.5K]

Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height.
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion).
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)?
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m

7 0

2 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago

The linear charge density on the inner conductor is and the linear charge density on the outer conductor is

Lubov Fominskaja [6]

Complete Question

The complete question is shown on the first uploaded image (reference for Photobucket )

Answer:

The electric field is $E = -1.3 *10^{-4} \ N/C$

Explanation:

From the question we are told that

The linear charge density on the inner conductor is $\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m$

The linear charge density on the outer conductor is

$\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m$

The position of interest is r = 37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

$E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}$

=> $E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}$

substituting values

$E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}$

$E = -1.3 *10^{-4} \ N/C$

The negative sign tell us that the direction of the electric field is radially inwards

=> $|E| = 1.3 *10^{-4} \ N/C$

5 0

3 years ago

1. What are electromagnetic waves?

Wewaii [24]

las ondas electromagnéticasSon aquellas ondas que no necesitan un medio material para propagarse. Incluyen, entre otras, la luz visible y las ondas de radio, televisión y telefonía. ... Las ondas electromagnéticas se propagan mediante una oscilación de campos eléctricos y magnéticos.

8 0

3 years ago