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3 years ago

How is the nitrogen cycle important to humans? HELP 18 mins

Physics

2 answers:

nirvana33 [79]3 years ago

6 0

Answer:

Its B

Explanation:

It converts nitrogen into a form that humans can obtain by eating other organisms.

nikdorinn [45]3 years ago

3 0

Nitrogen is a crucially important component for all life. It is an important part of many cells and processes such as amino acids, proteins and even our DNA. It is also needed to make chlorophyll in plants, which is used in photosynthesis to make their food.

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Fossil fuels store energy from the sun as<br><br> question 5

qwelly [4]

Answer:

Fossil fuels store energy from the sun as

Explanation:

3 0

3 years ago

What kind of soil is most likely found in the desert

Llana [10]

The most likely type of soil in a dessert is sand

3 0

3 years ago

1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential

cupoosta [38]

<h3>Answer:</h3>

$\displaystyle U_s = 0.6 \ J$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: $\displaystyle U_s = \frac{1}{2} k \triangle x^2$

U is energy (in J)
k is spring constant (in N/m)
Δx is displacement from equilibrium (in m)

<h3>Explanation:</h3>

<u>Step 1: Define</u>

k = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

Substitute in variables [Elastic Potential Energy]: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2$
Evaluate exponents: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = 0.6 \ J$

6 0

3 years ago

A block of mass 0.84 kg is suspended by a string which is wrapped so that it is at a radius of 0.061 m from the center of a pull

lora16 [44]

Answer:

$E_l = 1.713 J$

Explanation:

Given data:

mass of block is $M_b = 0.84 kg$

radius of block = 0.061 m

moment of inertia is $6.20 \times 10^{-3} kg m^2$

D is distance covered by block = 0.65 m

speed of block is 1.705 m/s

From conservation of momentum we have

$M_b g D = \frac{1}{2} M_b v^2 + \frac{1}{2} I \omega^2 + E_{loss}$

$0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times 0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l$

solving for energy loss

$E_l = 1.713 J$

3 0

3 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago