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neonofarm [45]
3 years ago
5

Which process of ammonia processing will Eden use? Eden owns a fertilizer manufacturing factory. Ammonia is used to prepare nitr

ogenous fertilizers in his factory. Eden will use , a popular process to prepare ammonia. NEED ANSWERSSS
Physics
2 answers:
Luda [366]3 years ago
5 0

Answer: Ammonia is critical in the manufacturing of fertilizers, and is one of the productions of significant quantities of ammonia using the cyanamide process. The plant used a four-case centrifugal compressor to compress the syngas to a Most modern plants can produce ammonia with an energy consumption of 28 GJ/m.t.

inessss [21]3 years ago
3 0

im sorry i dont know

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1. How much force would you have to apply to a 15kg object in order to accelerate it!<br> a 2 m/s?
mihalych1998 [28]

30N

Explanation:

Given parameters:

Mass of object = 15kg

Acceleration = 2m/s

Unknown:

Force = ?

Solution:

Force is given as the product of mass and acceleration:

         F = m x a

   m is the mass

    a is the acceleration

Inputting the parameters:

     F = 15 x 2 = 30N

The unit of force is newtons, N .

Learn more:

Force brainly.com/question/10470406

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3 years ago
If love is the answer, then what is the question?
kipiarov [429]

Answer:

what is happiness

Explanation:

7 0
3 years ago
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(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of
NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

where we subtract 180° because \vec F terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

8 0
2 years ago
What is the relationship between wavelength of light and the quantity of energy per photon?
slamgirl [31]

Answer:

The quantity of energy per photon is inversely proportional to the wavelength of the light.

Explanation:

Energy of light is given as

E = hf

where E = energy of the photons,

f = frequency of the light

If the number of photons = n

(E/n) = (h/n) f

Let (E/n) = E'

(h/n) = h'

But the frequency of light is related to wavelength through the relation

v = fλ

where v = speed of light = c

λ = wavelength of light

f = (c/λ)

E' = h' f

Substituting for f

E' = h' (c/λ)

h' and c are both constants, h'×c = K

E' = (K/λ)

So, the quantity of energy per photon is inversely proportional to the wavelength of the light.

Hope this Helps!!!

3 0
3 years ago
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9. A sailor pulls a boat along a dock using a rope at an angle of 60.0° with the
aleksklad [387]

Answer: (1) 3.83x10^3 J

Explanation:

(1) Fx=(255N)cos60°

   dx=30.0m

   w=Fx dx =(255)(cos60°)(30.0m)

6 0
3 years ago
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