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Zanzabum
3 years ago
5

The major horizontal surface on an airplane that provides lift.​

Physics
1 answer:
fgiga [73]3 years ago
5 0

wings

Lift is generated by every part of the airplane, but most of the lift on a normal airliner is generated by the wings. Lift is a mechanical aerodynamic force produced by the motion of the airplane through the air.

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What occurs when two Stars collide into each other?
ddd [48]

Answer:

A stellar collision.

Explanation:

A stellar collision is the coming together of two stars caused by stellar dynamics within a star cluster, or by the orbital decay of a binary star due to stellar mass loss or gravitational radiation, or by other mechanisms not yet well understood.

5 0
3 years ago
Read 2 more answers
Suppose you increase your walking speed from 5 m/s to 14 m/s in a period of 3 s. What is your acceleration
kirza4 [7]
5-14 m/s in 3 seconds
a=vf-vi/t
a=14-5/3
a=9/3
a=3 m/s^2
4 0
3 years ago
Will mark brainliest if its correct pls help
Nesterboy [21]

Answer:

I cant see the provided answer next to the options there is nothing

Explanation:

4 0
3 years ago
A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular
NeTakaya
Answer:
387 volts

Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)

In the given, we have:
current (I) = 9 amperes
resistance (R) = 43 ohm

Substitute with the givens in the above formula to get the voltage as follows:
V = 9 * 43
V = 387 volts

Hope this helps :)
4 0
3 years ago
Read 2 more answers
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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