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Nitella [24]
3 years ago
12

Am radio signals have frequencies between 550 khz and 1600 khz (kilohertz) and travel with a speed of 3.00 ✕ 108 m/s. what are t

he wavelengths of these signals?
Physics
1 answer:
Allisa [31]3 years ago
5 0
The minimum frequency is 
f_1 = 550 kHz = 5.50 \cdot 10^5 Hz
while the maximum frequency is
f_2 = 1600 kHz = 1.6 \cdot 10^6 Hz
Using the relationship between frequency f of a wave, wavelength \lambda and the speed of the wave v, we can find what wavelength these frequencies correspond to:
\lambda_1 =  \frac{v}{f_1}= \frac{3 \cdot 10^8 m/s}{5.5 \cdot 10^5 Hz}=545 m
\lambda_2 =  \frac{v}{f_2}= \frac{3 \cdot 10^8 m/s}{1.6 \cdot 10^6 Hz}=188 m

So, the wavelengths of the radio waves of the problem are within the range 188-545 m.

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The number of outer shell electrons determines the chemical properties of an element?
Ivan
Yes, it is true to a certain that the number of outer shell electrons determines the chemical properties of an element, only because this determines how the element interacts with other elements.
3 0
3 years ago
State advantages of ultrasonic sound in determining the depth of the ocean ​
agasfer [191]

Answer:

It has <u>greater accuracy than other nondestructive methods in determining the depth of internal flaws and the thickness of parts with parallel surfaces.</u>

Explanation:

Hope this helps you!

5 0
2 years ago
What value do we use to describe acceleration due to gravity
andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

7 0
3 years ago
A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b
musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E)  ⇒ P = ΔL*A*E/L

⇒  P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒  P = 24624000  N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒  σ = (24624000  N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax  we use σmax

⇒  P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

7 0
2 years ago
The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × 10^{8} Pa

Explanation:

The pressure at the depth is given by,

P = h \rho g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

\rho = density of water = 1000 \frac{kg}{m^{3} }

g = acceleration due to gravity ≈ 9.8 \frac{m}{s^{2} }

P = 11033.15 × 9.8 × 1000

P = 1.08 × 10^{8} Pa

Thus, the pressure at the depth is 1.08 × 10^{8} Pa

4 0
3 years ago
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