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3 years ago

PROPERTIES OF RADIO WAVES

Physics

2 answers:

Murrr4er [49]3 years ago

7 0

Answer:

Radio waves are a type of electromagnetic radiation with wavelengths in the electromagnetic spectrum longer than infrared light. Radio waves have frequencies as high as 300 gigahertz (GHz) to as low as 30 hertz (Hz).

Explanation:

vladimir2022 [97]3 years ago

5 0

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Bess [88]

Im not 100% sure you have to tell me if im wrong or not.

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3 years ago

Read 2 more answers

Find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1)to (3,-3). The amount of work done is ___?

dalvyx [7]

Answer:

amount of work done is $W = \frac{-4}{3}$

Explanation:

Formula for work done by force field

$W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt$

where

r(t) is parametrization of line

as it is straight line so

$r(t) = 1- t) r_o + tr_1 0 \leq t \leq 1$

thus,

r(t) = (1-t)(-1,1) + t(3,-3)

= (-1+t,1-t) + (3t - 3t)

= (-1+t +3t, 1-t-3t)

r(t) = (4t -1, 1- 4t)

r'(t) = (4,-4)

putting value in above integral

$\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \int_{0}^{1} (-(4t -1)^2 , 2-8t).(4,-4) dt$

$= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt$

$=\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt$

$= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}$

$\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \frac{-4}{3}$

3 0

4 years ago

10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)

Darya [45]

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .

Given: Weight = 475 N

$g= 9.8\ m/s^2$

Substitute all values in formula , we get

$475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg$

Hence, his mass = 48.47 kg.

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3 years ago

You venture out on a cold winter morning to warm up your vehicle. You have layers of cotton/polyester blend clothes on and from

xxMikexx [17]

Answer:

There is a localization of negative charge near the door handle.

4 0

3 years ago

A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

3 0

3 years ago

PROPERTIES OF RADIO WAVES​

PROPERTIES OF RADIO WAVES