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3 years ago

A radio wave has a frequency of 5.5 × 104 hertz and travels at a speed of 3.0 × 108 meters/second. What is its wavelength

Physics

2 answers:

gizmo_the_mogwai [7]3 years ago

6 0

Hope it helped you.

Ne4ueva [31]3 years ago

4 0

Use v=fλ
3x10^8=5.5x 10^4 λ
λ=5.45x10^3m

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How are land-use laws in the United States different from the land-use laws inherited from England?

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private property owners in the United States can restrict public access to their land

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3 years ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

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3 years ago

Neil pogo sticks to his science class, but stops to pick up his backpack on his way. He travels 8 m east, then 4 m west. what di

Likurg_2 [28]

Answer:

8.9 m NorthEast

Explanation:

a² + b² = c²

8² + 4² = c²

64 + 16 = c²

80 = c²

c = √80

c ≈ 8.9 m NorthEast

8 0

3 years ago

Pls helppp. Is this right?

andreyandreev [35.5K]

Answer:

yes you are totally right

7 0

3 years ago

Read 2 more answers

When you strike two tuning forks simultaneously, you hear three beats per second. The frequency of the first tuning fork is 440

tekilochka [14]

Answer:4

Explanation:

Given

Frequency of beats is 3 beats per second

Frequency of first tuning fork is $f_1=440\ Hz$

Beat frequency is difference in the frequency of tuning forks i.e.

either $f_1-f_2$ or $f_2-f_1 =3$

so $f_2=3+440=443\ Hz$

$f_2=440-3=437\ Hz$

so there is not enough information given to decide.

5 0

3 years ago

Read 2 more answers