The slope of the line on a speed-time graph tells the speed.

Mr. Adams asked his students to write the chemical symbol for the element zinc. He wrote the students’ answers on the whiteboar

m_a_m_a [10]

The chemical symbol for the element zinc is Zn.

<h3>What is element?</h3>

A chemical element is the one which reacts with other elements to form a compound.

Any chemical symbol must have its first letter written in Uppercase. If the element has two letter chemical symbol, then it should be written in lowercase.

Thus, the student wrote Zn as the chemical symbol of Zinc on the whiteboard.

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4 0

1 year ago

Is a reflection matter?

azamat

Answer:

Yes, because everything bounce off in every surface around any object.

Explanation:

7 0

3 years ago

Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l

ziro4ka [17]

Answer:

The pressure is 6570 lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

$a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s$

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

$M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3$

then, $M_n = M_1sin \beta = 3sin(35) = 1.721$

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570 lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

$M_t_1 = M_1cos \beta = 3 cos(35) = 2.458$

$w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s$

$a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s$

$v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s$

$Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7} \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o$

6 0

3 years ago

What is the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosp

sineoko [7]

The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

To find the answer, we have to know about the pressure.

<h3>How to find the weight of a column of air?</h3>

As we know that the expression of pressure as,

$P=\frac{F}{A}$

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.

It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

$P=1atm=1.013*10^5Pascals$

From this, the value of weight will be,

$F=mg=P*A=1.013*10^5*4.5=4.56*10^5N$

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

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4 0

1 year ago

Are the items in the list renewable or nonrenewable resources?

Elza [17]

N
R
R
R
N
Hope this helps.

8 0

3 years ago

Read 2 more answers