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Reika [66]
2 years ago
13

When an airplane experiences a tailwind, are the forces added or subtracted?

Physics
2 answers:
lorasvet [3.4K]2 years ago
8 0

Answer:

There is no force added or subtracted.

Explanation:

The aircraft moves within the air mass that it is part of and in order to compute the ground speed you would add the speed of the tailwind to the airspeed of the aircraft

ELEN [110]2 years ago
4 0
The answer is SUBTRACTED
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On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distanc
Arisa [49]

Answer:

Explanation:

The difference in time will be due to travel through atmosphere where speed of light slows down. If t be the thickness of atmosphere and c be the speed of light in space and μ be the refractive index of atmosphere difference in travel time will be as follows .

difference = \frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( 1-\mu  \right )

Now t = 40 x 10³m ,μ = 1.000293 , c = 3 x 10⁸.

difference =\frac{2t\mu }{c}-\frac{2t }{c}

=\frac{2t}{c }\left ( \mu -1  \right )\\

=\frac{ 2\times 40\times 10^3}{3\times10^3 }\left ( 1.000293-1 \right )\\

=7.81\times 10^{-3} s

4 0
4 years ago
a missile is moving 1810 m/s at a 20.0. It needs to hit a target 29,500m away in a 65.0 magnitude in 9.20s. What acceleration mu
Vesna [10]

Answer:

323 m/s²

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

x = 29500 cos 65°

y = 29500 sin 65°

v₀x = 1810 cos 20°

v₀y = 1810 sin 20°

t = 9.20

Find:

ax, ay, θ

First, in the x direction:

x = x₀ + v₀ t + ½ at²

29500 cos 32° = 0 + (1810 cos 20°) (9.20) + ½ ax (9.20)²

25017 = 15648 + 42.32 ax

ax ≈ 221.4

And in the y direction:

y = y₀ + v₀ t + ½ at²

29500 sin 32° = 0 + (1810 sin 20°) (9.20) + ½ ay (9.20)²

15633 = 5695 + 42.32 ay

ay ≈ 234.8

Therefore, the magnitude of the acceleration is:

a² = ax² + ay²

a² = (221.4)² + (234.8)²

a ≈ 322.7

Rounded to 3 significant figures, the magnitude of the acceleration is approximately 323 m/s².

7 0
3 years ago
A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00×105 Pa
Alborosie
<h2>The work done = - 2 x 10⁴ J</h2>

Explanation:

In the first case , the volume is kept constant and pressure varies .

In isothermal process  , the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume , therefore displacement is zero .

In second case pressure is constant , but volume changes

Thus W₂ = P x ΔV

here P is the pressure  and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative .

7 0
4 years ago
A camera phone called the iPhone X has an image sensor size of 5 mm and a focal length of 4 mm. How long of a selfie-stick must
azamat

Answer:

length of selfie-stick is 1.62 m

Explanation:

Given data

image size h1 = 5 mm = 5 ×10^{-3} m

focal length = 4 mm = 4 ×10^{-3} m

distance h2 = 2.032 m

to find out

How long of a selfie-stick

solution

here we find first magnification

that is M = h1 /h2

M = 5 ×10^{-3} / 2.032

M = 2.46 ×10^{-3}

and we know M = p/q

so p = Mq = 2.46 ×10^{-3} q

so we apply lens formula

1/f = 1/p - 1/q

1/ 4 ×10^{-3} = 1 / 2.46 ×10^{-3}  q - 1/q

q = 1.622 m

so length of selfie-stick is 1.62 m

4 0
3 years ago
Bromine vapour is heavier than air. even so it's spreads upwards in the experiment above. Why?
Yakvenalex [24]

Answer:

hi how are u fine thanks you

4 0
3 years ago
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