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The FM radio band in most places goes from frequencies of about 89 MHz to 106 MHz. How long is the wavelength of the radiation a

ohaa [14]

Answer:

2.83 m

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is given by

$\lambda=\frac{c}{f}$

where

$\lambda$ is the wavelength

$c=3.0\cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

For the FM radio waves in this problem, we have:

$f_1=89 MHz=89\cdot 10^6 Hz$ is the minimum frequency, so the maximum wavelength is

$\lambda_2=\frac{c}{f_1}=\frac{3\cdot 10^8}{89\cdot 10^6}=3.37 m$

The maximum frequency is instead

$f_2=106 MHz=106\cdot 10^6 Hz$

Therefore, the minimum wavelength is

$\lambda_1=\frac{c}{f_2}=\frac{3\cdot 10^8}{106\cdot 10^6}=2.83 m$

So, the wavelength at the beginning of the range is 2.83 m.

8 0

3 years ago

In carbon dioxide (CO2), there are two oxygen atoms for each carbon atom. Each oxygen atom forms a double bond with carbon, so t

Anton [14]

ANSWER: d) 8

EXPLANATION: Two sets of two shared electrons (4 electrons total shared) = one set of a double covalent bond.

Therefore, 8 electrons total shared = two sets of double covalent bonds

5 0

3 years ago

A cyclist is taking part in the Tour de France, which is a bicycle race that takes place every year.

marishachu [46]

Solve the following word problems.
1. The ratio of red marbles and blue marbles that Carlo has is 8: 3. When he
exchanged 35 red marbles for 20 blue marbles from his brother, he was left with
equal number of red and blue marbles.
How many red and blue marbles did he have at the beginning
How many red and blue marbles did he have now

5 0

3 years ago

Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer dista

nadezda [96]

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s) = 344 meters

Car B: Distance = (7 m/s) x (50 s) = 350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

5 0

3 years ago

Read 2 more answers

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

2.78m/s²

Explanation:

Complete question:

A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

Hence the acceleration of the box as it slides down the incline is 2.78m/s²

5 0

3 years ago