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2 years ago

Which of the choices below is one of the primary gases found in the atmosphere? A. Helium B. Carbon Dioxide C. Nitrogen D. Argon

Physics

2 answers:

Goshia [24]2 years ago

7 0

C. Nitrogen is a primary gas found in the atmosphere

sp2606 [1]2 years ago

3 0

Answer:

Nitrogen - 78%

Explanation:

Our atmosphere is a combination of many different types of gases. The gases that are present in our atmosphere are as follows :

(1) Oxygen - 21 %

(2) Nitrogen - 78 %

(3) Carbon dioxide - 0.04 %

(4) Argon - 0.93 %

The amount of other gases like neon, methane hydrogen etc are very small.

Hence, the primary gases that found in the atmosphere is (c) " nitrogen ".

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3. Tightly Packed

Explanation:

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2 years ago

An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y

Xelga [282]

Answer:

The fraction of the protons would have no electrons $=1.88\times 10^{-10}$

Explanation:

We are given that

Amoeba has total number of protons= $1.00\times 10^{16}$

Net charge, Q=0.300pC

Electrons are fewer than protons= $1.88\times 10^6$

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

= $\frac{Fewer\;electrons}{Total\;protons}$

The fraction of the protons would have no electrons

= $\frac{1.88\times 10^{6}}{1.00\times 10^{16}}$

$=1.88\times 10^{-10}$

Hence, the fraction of the protons would have no electrons $=1.88\times 10^{-10}$

6 0

2 years ago

A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

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2 years ago

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goblinko [34]

Sound waves and light waves need to change

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