All categories

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.

You might be interested in

Ayo, how do I change my username on here?

nydimaria [60]

Answer:

I'm not sure

Explanation:

eeeeeeeeeeeeeeeeeeeeeee

4 0

3 years ago

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) a

Fudgin [204]

Answer:

5.21e-2mm

Explanation:

Please see attachment

8 0

3 years ago

If you are trying to land the plane, which part(s) would you move? How would you move it/them?

KiRa [710]

Answer: hope it helps

Explanation:Moving air has a force that will lift kites and balloons up and down. Air is a mixture ... Here is a simple computer simulation that you can use to explore how wings make lift. ... All these dimensions together combine to control the flight of the plane. A pilot ... When the rudder is turned to one side, the airplane moves left or right.

7 0

3 years ago

Two balls are chosen randomly from an urn containing 8 white 4 black, and orange balls. Suppose that we win $ 2 for each black b

Scorpion4ik [409]

(-2,-10,-1,-2,-3,-4)

8 0

3 years ago

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

liberstina [14]

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

i (t) = 6*e^(-2*t)

v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

- The amount of charge Q delivered can be determined by:

dQ = i(t) . dt

$Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt$

- Integrate and evaluate the on the interval:

$= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

- The power can be calculated by using v(t) and i(t) as follows:

v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

$W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt$

- Integrate and evaluate the on the interval:

$W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ$

6 0

3 years ago