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GuDViN [60]
3 years ago
11

If the surface air pressure is 1000 mb and the pressure at the top of the atmosphere (75 km) is 0 mb, at what altitude would I f

ind half of the atmosphere air pressure?
Physics
1 answer:
Lana71 [14]3 years ago
8 0

Answer: 5.5km

Explanation:

Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).

Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.

Pressure decreasing with higher altitudes also means that  air pressure decreases rapidly at lowerevels but more slowly at higher levels.

It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).

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The matter from the explosion can reach him, hitting him. He should be able to feel that.
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Which statement describes the effects of forces on an object
lukranit [14]

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

8 0
3 years ago
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give
Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

Explanation:

From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

4 0
3 years ago
What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copp
Lina20 [59]

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

Q=mC\Delta T

where:

m is the mass of the substance

C is the specific heat capacity of the substance

\Delta T is the change in temperature of the substance

For the sample of copper in this problem, we have:

m = 25 g (mass)

C = 0.39 J/gºC (specific heat capacity of copper)

\Delta T = 15^{\circ}C (change in temperature)

Substituting, we find:

Q=(25)(0.39)(15)=146 J

So, the closest answer is B) 150 J.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
A light wave has a wavelength of 450 nanometers. What is the frequency of this light?
Zolol [24]

Answer:

Frequency, f=6.67\times 10^{14}\ Hz

Explanation:

Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,

c=f\lambda

f is the frequency of this light

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{450\times 10^{-9}}\\\\f=6.67\times 10^{14}\ Hz

So, the frequency of this light is 6.67\times 10^{14}\ Hz.

3 0
3 years ago
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