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2 years ago

What Is Photosynthesis ?

Engineering

2 answers:

Fynjy0 [20]2 years ago

6 0

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

Juliette [100K]2 years ago

4 0

Answer:

Photosynthesis is the process by which plants use sunlight, water, and carbon dioxide to create oxygen and energy in the form of sugar.

have a nice day <3

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Air at 80 kPa and 10Â°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

$A_2=5A_1$

If consider that density of air is not changing from inlet to exit then by using continuity equation

$A_1V_1=A_2V_2$

So $A_1\times 150=5A_1V_2$

$V_2=30$ m/s

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here Q=0 and w=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

When air is treating as ideal gas

$h=C_pT$

Noe by putting the values

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

$1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}$

$T_2=293.74K$

So the exit temperature is 293.74 K.

7 0

3 years ago

What’s the number of gold atoms in a nanogram? a picogram?

zvonat [6]

Answer :

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

The number of gold atoms in picogram is, $3.057\times 10^{9}$

Explanation :

As we know that the molar mass of gold is, 196.97 g/mole. That means, 1 mole of gold has 196.97 grams of mass of gold.

As we know that,

1 mole contains $6.022\times 10^{23}$ number of atoms.

First we have to determine the number of gold atoms in a nanogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-9}$ nanograms of gold contains $\frac{1g}{196.97g}\times (10^{-9})\times (6.022\times 10^{23})=3.057\times 10^{12}$ number of gold atoms

The number of gold atoms in nanogram is, $3.057\times 10^{12}$

Now we have to determine the number of gold atoms in a picogram.

As, 196.97 grams of gold contains $6.022\times 10^{23}$ number of gold atoms

And, 1 grams of gold contains $\frac{1g}{196.97g}\times (6.022\times 10^{23})$ number of gold atoms

So, $10^{-12}$ picograms of gold contains $\frac{1g}{196.97g}\times (10^{-12})\times (6.022\times 10^{23})=3.057\times 10^{9}$ number of gold atoms

The number of gold atoms in picogram is, $3.057\times 10^{9}$

8 0

3 years ago

What is the formula used to find the volume of this shape

horsena [70]

Volume=Hh(b1+b2)/2. b1 and b2 are the base of the trapezoid

8 0

3 years ago

A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball

faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

$\sigma = \frac{p*r}{2t}$

Where:

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm

Hence, we need to find r, as follows:

$r_{inner} = r_{outer} - t$

$r_{inner} = \frac{d}{2} - t$

Where:

d: is the outer diameter = 300 mm

$r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm$

Now, we can find the normal stress (σ) in the wall of the basketball:

$\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa$

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0

3 years ago

A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 s

Nezavi [6.7K]

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) = 2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

6 0

3 years ago

What Is Photosynthesis ?​

What Is Photosynthesis ?