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LuckyWell [14K]

3 years ago

5

Rubber bands provided the thrust to get your rocket going. Why did the rocket keep moving after it left the launch pad and the r

ubber bands were no longer pushing it?

1 answer:

BARSIC [14]3 years ago

3 0

Answer:

rocket engines

Explanation:

You might be interested in

Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to

7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

= 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

= 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0

3 years ago

Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol

sleet_krkn [62]

Answer:

$Q=15.7Kw$

Explanation:

From the question we are told that:

Initial Pressure $P_1=4bar$

Initial Temperature $T_1=20 C$

Final Pressure $P_2=12 bar$

Final Temperature $T_2=80C$

Work Output $W= 60 kJ/kg$

Generally Specific Energy from table is

At initial state

$P_1=4bar \& T_1=20 C$

$E_1=262.96KJ/Kg$

With

Specific Volume $V'=0.05397m^3/kg$

At Final state

$P_2=12 bar \& P_2=80C$

$E_1=310.24KJ/Kg$

Generally the equation for The Process is mathematically given by

$m_1E_1+w=m_2E_2+Q$

Assuming Mass to be Equal

$m_1=m_1$

Where

$m=\frac{V}{V'}$

$m=frac{0.06666}{V'=0.05397m^3/kg}$

$m=1.24$

Therefore

$1.24*262.96+60)=1.24*310.24+Q$

$Q=15.7Kw$

4 0

3 years ago

The dot plot shows the number of cupcakes bought by each person who came to a bake sale.

Maslowich

The true statement about the dot plot is 1 has 4 and 0 dots.

Explanation:

after creating a Bar chart 4 is the right answer.
The highest among st all the other plots is 1 but 4 shows 3.
Taking an average from all the data points 3 comes to the right answer.
Median the central Mid-point is 3.
Mode also comes to 3.
It is skewed on the right due to the 1st one.
Skewed shows the data point either increasing or in decreasing.
There a to Bi- histograms which has two ups and downs.
The true statement has to be as Mean=Median=Mode is 3.

6 0

3 years ago

what is called periodic function give example? Plot the output which is started with zero degree for one coil rotating in the un

marta [7]

Answer:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin <em>wt</em>

Explanation:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin <em>wt</em>

attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field

B ( wave flux density ) = Bm sin<em>wt and w = </em>2 $\pi$ f = $\frac{2\pi }{T}$ rad/sec

3 0

3 years ago

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=cons

antoniya [11.8K]

Answer:

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Explanation:

Given :

$PV^{1.3}=C$

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

$\eta = 1-\left ( \frac{1}{r_{c}} \right )^{1.3-1}$

where $r_{c}$ is compression ratio = $\frac{v_{c}+v_{s}}{v_{c}}$

$r_{c}$ = $1+\frac{v_{s}}{v_{c}}$

where $v_{c}$ is compression volume

$v_{s}$ is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then, $v_{30}=v_{c}+0.7v_{s}$

and at 70% compression, 30% of the swept volume remains

∴ $v_{70}=v_{c}+0.3v_{s}$

We know,

$\frac{P_{2}}{P_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{n}$

$\frac{2.75}{1.5}=\left ( \frac{v_{c}+0.7\times v_{s}}{v_{c}+0.3\times v_{s}} \right )^{1.3}$

$\left ( 1.833 \right )^{\frac{1}{1.3}}=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}\\$

$1.594=\frac{v_{c}+0.7v_{s}}{v_{c}+0.3v_{s}}$

$v_{c}+0.7v_{s}=1.594v_{c}+0.4782v_{s}$

$0.7v_{s}-0.4782v_{s}=1.594v_{c}-v_{c}$

$0.2218v_{s} = 0.594v_{c}$

$v_{c}=0.3734 v_{s}$

∴ $r_{c}= 1+\frac{v_{s}}{0.3734v_{s}}$

Therefore, compression ratio is $r_{c}$ = 3.678

Now efficiency, $\eta =\left ( 1-\frac{1}{r_{c}} \right )^{0.3}$

$\eta =\left ( 1-\frac{1}{3.678} \right )^{0.3}$

$\eta =0.32342$ , this is the ideal efficiency

Therefore actual efficiency, $\eta_{act} =0.5\times \eta _{ideal}$

$\eta_{act} =0.5\times 0.32342$

$\eta_{act} =0.1617$

Therefore total power required = 1 kW x 3600 J

= 3600 kJ

∴ we know efficiency, $\eta=\frac{W_{net}}{Q_{supply}}$

$Q_{supply}=\frac{W_{net}}{\eta _{act}}$

$Q_{supply}=\frac{3600}{0.1617}$

$Q_{supply}=22261.78 kJ$

Therefore fuel required = $\frac{22261.78}{45000}$

= 0.4947 kg/hr

5 0

4 years ago