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3 years ago

What part of the airplane helps to provide it with lift?

Physics

1 answer:

Dmitrij [34]3 years ago

5 0

Answer:

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The elevators which are on the tail section are used to control the pitch of the plane. A pilot uses a control wheel to raise and lower the elevators, by moving it forward to back ward. Lowering the elevators makes the plane nose go down and allows the plane to go down.

Explanation:

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What changes in airplane longitudinal control must be made to maintain altitude while the airspeed is being decreased?

garik1379 [7]

Explanation:

The changes can be made in airplane longitudinal control to maintain altitude while the airspeed is being decreased is

We can increase the angle of attack this would compensate for the decreasing lift. As the angle of attack directly controls the distribution of pressure on the wings. Moreover, increase in angle of attack will also cause the drag to increase.

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3 years ago

How does the electric force between two charged participles change if one particle’s charge is reduced by a factor of 3?

Eva8 [605]

Answer:

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Explanation:

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3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago

PLEASE HELP I need the answer to this question ASAP

cestrela7 [59]

Answer:

answer is A

Explanation:

hope this helps!

7 0

2 years ago

Read 2 more answers

A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V lin

bulgar [2K]

Back emf is 85.9 V.

<u>Explanation:</u>

Given-

Resistance, R = 3.75Ω

Current, I = 9.1 A

Supply Voltage, V = 120 V

Back emf = ?

Assumption - There is no effects of inductance.

A motor will have a back emf that opposes the supply voltage, as the motor speeds up the back emf increases and has the effect that the difference between the supply voltage and the back emf is what causes the current to flow through the armature resistance.

So if 9.1 A flows through the resistance of 3.75Ω then by Ohms law,

The voltage across the resistance would be

v = I x R

= 9.1 x 3.75

= 34.125 volts

We know,

supply voltage = back emf + voltage across the resistance

By plugging in the values,

120 V = back emf + 34.125 V

Back emf = 120 - 34.125

= 85.9 Volts

Therefore, back emf is 85.9 V.

4 0

3 years ago