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allochka39001 [22]
3 years ago
11

How far above Earth's surface do Gamma Waves reach?

Physics
2 answers:
velikii [3]3 years ago
8 0
100 miles from earth

lapo4ka [179]3 years ago
8 0
I am pretty sure around 100 miles from Earth
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A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radiu
anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

7 0
4 years ago
The cylinder valve is open and the gas is collected and atmospheric pressure
jok3333 [9.3K]

Less gas will be collected because some of the gases will escape from the open cylinder valve.

Cylinders used to store carbon dioxide will have thicker walls than those of butane because of higher pressures.

<h3>What are compressed gases?</h3>

Compressed gases are gases which are compressed under high pressure in gas cylinders.

Cylinder valves are used to reduce the pressure of the compressed gases and in the process, some of the gas molecules escape.

Since the cylinder valve is open and the gas is collected at atmospheric pressure, less gas will be collected because some of the gases will escape.

Since, the carbon dioxide not liquefy under pressure compared to butane, the cylinders used to store carbon dioxide will have thicker walls than those of butane.

Learn more about compressed gases at: brainly.com/question/518065

4 0
2 years ago
How large a band of frequencies does each television broadcasting channel get ?
k0ka [10]

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

8 0
3 years ago
Read 2 more answers
Which of the following are early clues that signal an earthquake may occur? Select all that apply. Changes in magnetic propertie
enot [183]

The following answers apply;

  • Changes in magnetic properties of rock
  • Decrease in well water levels
  • Increases in radon gas in groundwater
  • Foreshocks  

These other choices may be good indicators of an imminent volcanic eruption;

Movement of magma

Increase in sulfur dioxide and carbon dioxide ground emissions


6 0
3 years ago
Read 2 more answers
Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of
Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m  

we know,

c = f × λ  

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency  

thus,

f=\frac{c}{\lambda}

substituting the values in the equation we get,

f=\frac{3.0\times 10^8 m/s}{2.9m}

f = 1.03 x 10⁸Hz  

Now,

The time period (T) = \frac{1}{f}

or

T =  \frac{1}{1.03\times 10^8}  = 9.6 x 10⁻⁹ seconds  

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s  

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s  

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}  

Thus, the minimum distance to target = \frac{290}{2} = 145 m

6 0
3 years ago
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