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wolverine [178]
3 years ago
6

How and why does air pressure change with altitude in the atmosphere?

Physics
1 answer:
wlad13 [49]3 years ago
5 0
Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.
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When an object is lifted against the force of gravity it has gravitational potential energy. The energy depends on the object's
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The answer is actually True, I just took the test and it was correct.
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EXPLAIN CONSERVATION OF MOMEMTUM WITH THE HELP OF COLLISION OF TWO OBJECT.
Vlada [557]

i hate this question.

Conservation of momentum - is when the total momentum before and after collision is equal.

Here is the formula darling,

p = p

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3 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
3 years ago
Two long, straight parallel wires are placed 38 cm apart, one above the other. The top and bottom wires are carrying currents 4.
ElenaW [278]

Answer:

The force per unit length (N/m) on the top wire is 16.842 N/m

Explanation:

Given;

distance between the two parallel wire, d = 38 cm = 0.38 m

current in the first wire, I₁ = 4.0 kA

current in the second wire, I₂ = 8.0 kA

Force per unit length, between two parallel wires is given as;

\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d }

where;

μ₀ is constant = 4π x 10⁻⁷ T.m/A

Substitute the given values in the above equation and calculate the force per unit length

\frac{F}{L} = \frac{\mu_oI_1I_2 }{2\pi d } = \frac{4\pi *10^{-7}*4000*8000 }{2\pi *0.38} = 16.842 \ N/m

Therefore, the force per unit length (N/m) on the top wire is 16.842 N/m

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Why couldnt mendeleev organize the entire table during his research
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Cause he left out the noble gases out of the periodic table for one good reason, 1: He did not know them
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