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3 years ago

5. If a pilot wants to accelerate, which force

Engineering

1 answer:

Murljashka [212]3 years ago

3 0

Answer:

Thrust

Explanation:

If the pilot wants to accelerate the aircraft, the aircraft needs more power to produce more thrust. The aircraft will go faster when the amount of thrust is greater than the amount of drag.

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How far away must a pwc stay from anyone being towed behind another vessel?.

julia-pushkina [17]

Answer:

5 lengths away or 50-100 feet

Explanation:

I would say and from what I found 5 lengths away or 50-100 feet

7 0

2 years ago

For this problem, calculate the following by hand and show the procedure for how you obtained the results. Subsequently, solve p

Gemiola [76]

Answer:wat

Explanation:

4 0

3 years ago

Describe SIX basic ways of installing wire ways

shutvik [7]

1. Cleat wiring
2. Batten wiring
3. Conduit wiring
4. Casing and capping wiring
5. Lay in
6.Cable/Pull through

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3 years ago

In an oscillating LC circuit, L ! 25.0 mH and C ! 7.80 mF. At time t 0 the current is 9.20 mA, the charge on the capacitor is 3.

topjm [15]

Answer:

a) 926 μJ

b) 3.802 mC

c) 8.61 A

d) 0.0721

e) 3.2137

Explanation:

The energy in the inductor is

$El = \frac{1}{2}*L*I^2$

$El = \frac{1}{2}*25*10^{-3}*(9.2*10^{-3})^2 = 1.06*10^{-6} J = 1.06 \mu J$

The energy store in a capacitor is

$Ec = \frac{1}{2}*C*V^2$

The voltage in a capacitor is

V = Q/C

$V = \frac{3.8*10^{-3}}{7.8*10^{-3}} = 0.487 V$

Therefore:

$Ec = \frac{1}{2}*7.8*10^{-3}*0.487^2 = 9.256*10^{-4} J = 925.6 \mu J$

The total energy is Et = 925.6 + 1.1 = 926.7 μJ

At a certain point all the energy of the circuit will be in the capacitor, at this point it will have maximum charge

$Ec = \frac{1}{2}*C*V^2$

V = Q/C

$Ec = \frac{1}{2}*C*(\frac{Q}{C})^2$

$Ec = \frac{1}{2}*\frac{Q^2}{C}$

$Q^2 = 2*Ec*C$

$Q = \sqrt{2*Ec*C}$

$Q = \sqrt{2*926*10{-6}*7.8*10^{-3}} = 3.802 * 10{-3} C = 3.802 mC$

When the capacitor is completely empty all the energy will be in the inductor and current will be maximum

$El = \frac{1}{2}*L*I^2$

$I^2 = 2*\frac{El}{L}$

$I = \sqrt{2*\frac{El}{L}}$

$I = \sqrt{2*\frac{926.7*10^{-3}}{25*10^{-3}}} = 8.61 A$

At t = 0 the capacitor has a charge of 3.8 mC, the maximum charge is 3.81 mC

q = Q * cos(vt + f)

q(0) = Q * cos(v*0 + f)

3.8 = 3.81 * cos(f)

cos(f) = 3.8/3.81

f = arccos(3.8/3.81) = 0.0721

If the capacitor is discharging it is a half cycle away, so f' = f + π = 3.2137

4 0

3 years ago

A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass

jeyben [28]

Answer:

2.728 kg

Explanation:

The units help you keep the calculation straight.

$\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}$

4 0

3 years ago