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spayn [35]
3 years ago
6

Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons

ider a drag of 12 kN on the shuttle. Calculate the work done by the shuttle engine.
Engineering
1 answer:
mixas84 [53]3 years ago
6 0

Answer:

work done = 48.88 × 10^{9} J

Explanation:

given data

mass = 100 kN

velocity =  310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time  - force ×velocity ×time  - mass ×descends )  10³ J

work done = ( 100 × 310 × 1800  - 12×310 ×1800  - 100 × 2200 )  10³ J

work done = 48.88 × 10^{9} J

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Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series  circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

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3 years ago
If they opened up the International Space Station to tourism, would you go? Why? answer in 2 sentences
Arlecino [84]
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6 0
3 years ago
A smooth concrete pipe (1.5-ft diameter) carries water from a reservoir to an industrial treatment plant 1 mile away and dischar
Kamila [148]

ANSWER:

Q = 0.17ft3/s

EXPLANATION: since the water runs downhill on a 1:100 slope, that means the flow is laminar.

Using poiseuille equation:

Q = (π × D^4 × ∆P) ÷ (128 × U × ∆X)

Q is the volume flow rate.

π is pie constant value at 3.142

D is the diameter of the pipe

∆P is the pressure drop

U is the viscosity

∆X is the length of the pipe or distance of flow.

Form the question, we are to determine U then Find Q

Therefore;

D = 1.5ft

∆P = 1pa since the minor losses are negligible.

∆X = 1mile = 5280ft.

STEP1: FIND U

Viscosity is a function of the temperature of the liquid. An increase in temperature increases the viscosity of the liquid.

We know that at room temperature, which is 25°C the viscosity of water is 8.9×10^-4pa.s . We can find the viscosity of water at 4°C by cross multiplying.

Therefore;

25°C = 8.9×10^-4pa.s

4°C = U

Cross multiply

U25°C = 4°C × 8.9×10^-4pa.s

U25°C = 0.00356°C.pa.s

Therefore;

U = 0.00356°C.pa.s ÷ 25°C

U = 1.424×10^-4pa.s

Therefore at 4°C the viscosity of water in the pipe is 1.424×10^-4pa.s

STEP2: FIND Q

Imputing the values into poiseuille equation above.

Q = (3.142 × (1.5ft)^4 × 1pa) ÷ (128 × 1.424×10^-4pa.s × 5280ft)

Q = 15.906375pa.ft4 ÷ 96.239616pa.s.ft

Therefore;

Q = 0.16547887ft3/s

Approximately;

Q = 0.17ft3/s

6 0
3 years ago
Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes th
Svet_ta [14]

Answer:

The time required to elute the two species is 53.3727 min

Explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621

The mathematical expression for the time to elute the two species is:

\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}

Here

ReB = 1.5

t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min

6 0
2 years ago
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