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Bumek [7]
2 years ago
15

What is the volicity of a rocket?

Engineering
2 answers:
Marysya12 [62]2 years ago
8 0

Answer:

7.9 kilometers per second

Explanation:

Bas_tet [7]2 years ago
7 0

Answer:

I'm gonna assume you meant velocity.

Explanation:

If a rocket is launched from the surface of the Earth, it needs to reach a speed of at least 7.9 kilometers per second (4.9 miles per second) in order to reach space. This speed of 7.9 kilometers per second is known as the orbital velocity, it corresponds to more than 20 times the speed of sound.

You might be interested in
. An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to
tatuchka [14]

Answer:

Actual COP = 5.368

Maximum theoretical COP = 6.368

Explanation:

Given - An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC.

To find -  If this device operates using R134a as the working fluid. Calculate the actual COP of this device as well as the maximum theoretical COP.

Proof -

Given that,

An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa.

From Refrigerant 134-a Table

At T1 = -5°C

h1 = 247.505 KJ/kg

S1 = 0.93434 KJ/kg

At P2 = 900 KPa

S1 = S2

h2 = 274.679 Kj/Kg

h3 = h4 = 101.61 KJ/g

So,

Compressor work (Wc) = h2 - h1

                                       = 274.679 - 247.505

                                       = 27.174

⇒Compressor work (Wc) = 27.174 KJ/kg

Now,

Heat out (Qout) = h2 - h3

                          = 274.679 - 101.61

                          = 173.069

⇒Heat out (Qout) = 173.069 KJ/kg

Now,

Heat input (Qin) = h1 - h4

                          = 274.505 - 101.61

                          = 145.895

⇒Heat input (Qin) = 145.895 KJ/kg

So,

Actual COP at the refrigerator is -

(COP)R = (Qin)/(Wc)

            = (145.895)/ (27.174)

            = 5.368

⇒Actual COP = 5.368

Now,

Maximum theoretical COP is -

(COP) = (Qout)/(Wc)

          = (173.069)/ (27.174)

          = 6.368

⇒Maximum theoretical COP = 6.368

8 0
3 years ago
The reverse water-gas shift (RWGS) reaction is an equimolar reaction between CO2 and H2 to form CO and H2O. Assume CO2 associati
klasskru [66]

Answer:

a) Check explanation for this

b)Rate law is  Rate = \frac{k_{1}k_{4}  }{k_{3}+ 2k_{4}  } [H_{2} ]

c) The rate does not depend on the concentration of CO₂

Explanation:

a) Elementary steps for the RWGS reaction:

  • Dissociative adsorption of the H₂ Molecule

                 H_{2} $\xrightarrow{\text{k1}}$H + H   (Fast process)

  • Reversible Reaction between CO₂ and H

                \[ CO_{2} + H\mathrel{\mathop{\rightleftarrows}^{\mathrm{k2}}_{\mathrm{k3}}}COOH \] (Fast Process)

  • Slow dissociation of COOH into gaseous CO and absorbed OH

                COOH $\xrightarrow{\text{k1}}$ CO + OH (Slow process)

  • Fast hydrogenation of the OH to form H₂O

                   OH + H $\xrightarrow{\text{k5}}$H_{2} O (Fast process)

b) Derivation of the rate law

We need to determine the rate law for H, OH and COOH because these are the intermediates for this reaction.

The steady state approximation is applied to a consecutive reaction with a slow first step and a fast second step (k1≪k2). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product.

Rate of consumption = Rate of production

For COOH:

Using steady state approximation

\frac{d[COOH]}{dt} = 0

k_{2} [CO_{2} ][H] = k_{3} [COOH] k_{4} [COOH]\\

[COOH] = \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4}  } \\

For H:

\frac{d[H]}{dt} = 0

k_{1}[H_{2}] = k_{2}[CO_{2} [H]+k_{5} [ OH][H]

[H]= \frac{k_{1}[H_{2}]  }{k_{5}[OH] +k_{2}[CO_{2}]}\\

For OH:

\frac{d[OH]}{dt} = 0

k_{4} [COOH] = k_{5} [OH][H]\\\k[OH] = \frac{k_{4} [COOH]}{k_{5} H}\\

The rate of the overall reaction is determined by the slowest step of the reaction. The slowest process is the dissociation of COOH

Therefore the overall rate of reaction is:

Rate = k_{4} [COOH]\\

Rate = k_{4}  \frac{k_{2} [CO_{2} ][H]}{k_{3}k_{4}  }\\Rate = k_{4}  \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}]  }{k_{5}[OH] +k_{2}[CO_{2}]}  }{k_{3}k_{4}}\\Rate = k_{4}  \frac{k_{2}[CO_{2}]\frac{k_{1}[H_{2}]  }{k_{5}\frac{k_{4}COOH }{k_{5}H }  +k_{2}[CO_{2}]}  }{k_{3}k_{4}}

Simplifying the equation above, the rate law becomes

Rate = \frac{k_{1}k_{4}  }{k_{3}+ 2k_{4}  } [H_{2} ]

c) It is obvious from the rate law written above that the rate of the RWBG reaction does not depend on the concentration of CO₂

7 0
3 years ago
The component of a fluid system where a fluid is stored, but not under pressure, is called a container.
forsale [732]

Answer:

The answer is true

please friend me thank you bye.

Explanation:

6 0
3 years ago
I'm bored. I want to talk to you humans! weird. STUPID QUARANTINE!!! So. What are yall up to?
kiruha [24]

Answer:

nothing much what class r u in

3 0
3 years ago
A commercial refrigerator with refrigerant -134a as the working fluid is used to keep the refrigerated space at -30C by rejectin
Mariana [72]

Answer:

a) 0.487

b) refrigeration load = 5.46w

c) cop = 2.24

d)ref load max = 12.43kw

Explanation:

6 0
3 years ago
Read 3 more answers
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