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olga_2 [115]
3 years ago
11

An aeroplane cruises at normal speed. Upon fl ying into turbulent (unstable) air, it suddenly drops downwards vertically. Passen

gers in the aeroplane who are not wearing seat belts will A hit the seats in front of them. B hit the ceiling of the aeroplane. C be pushed against the back of their seats. D be pushed down into their seats.
Physics
1 answer:
fredd [130]3 years ago
6 0

Answer:

Your answer would be D. be pushed down into their seats.

Explanation:

You can think of it this way:

If you're not wearing a seat belt on an airplane that drops suddenly, in this case, vertically, which often happens with turbulence- you're the one at rest. You'll stay at rest as the plane (literally) drops out from under you.

If you're strapped in, the seat belt serves as an outside force acting on you, taking you with the plane as it drops and saving you from hitting the ceiling.

Always remember Newton's first law of motion: A body at rest will remain at rest unless an outside force acts on it.

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The volume of the lung 0.0024m^3 following exhalation and the pressure is 101.70KPa. Calculate the volume of the lungs during in
Mekhanik [1.2K]

According to Boyle-Mariotte:

p₁V₁=p₂V₂=>V₂=p₁V₁/p₂= 0.0024*101.70/ 84.16=0.0028 m³

5 0
4 years ago
21
sergeinik [125]
Distance between the centers of the two objects are the second object
5 0
2 years ago
A plane flying against the wind covers the 900-kilometer distance between two aerodromes in 2 hours. The same plane flying with
Mashcka [7]

Answer:

The speed of the wind is 25 km/hr.

Explanation:

Let us call v_p the speed of the plane and v_w the speed of the wind. When the plane is flying against the wind, it covers the distance of 900-km in 2 hours (120 minutes); therefore;

(1). v_p - v_w = \dfrac{900km}{120min}

And when the plane is flying with the wind, it covers the same distance in 1 hour 48 minutes (108 minutes)

(2). v_p+v_w= \dfrac{900km}{108min}

From equation (1) we solve for v_p and get:

v_p = \dfrac{900km}{120min}+v_w,

and by putting this into equation (2) we get:

\dfrac{900km}{120min}+v_w+v_w= \dfrac{900km}{108min}

2v_w= \dfrac{900km}{108min}-\dfrac{900km}{120min}

2v_w = 8.3km/min - 7.5km/min

2v_w = 0.83km/min

v_w = 0.4165km/min

or in km/hr this is

\boxed{v_w= 25km/hr }

4 0
3 years ago
Using the information from Paul Hewitt's Conceptual Development Practice Page 25-1 and the image below, answer the following que
MaRussiya [10]

Answer:

A = 2 cm ,   λ = 8 cm

Explanation:

The amplitude of a wave is the maximum height it has, in this case the height is measured by the vertical ruler,

We are told the balance point is in the reading of 5 cm, that the maximum reading is 3 cm and the Minimum reading is 7 cm. Therefore, the distance from the ends of the ridge to the point of equilibrium is

          d = 7-5 = 2 cm

          d = 5-3 = 2 cm

          A = 2 cm

The wavelength is the minimum horizontal distance for which the wave is repeated, that is measured by the horizontal ruler.

The initial reading for 4 cm and the final reading for 8 cm, this distance corresponds to a crest of the wave, the complete wave is formed by two crests whereby the wavelength is twice this value

          Δx = 8-4 = 4 cm

          λ = 2 Δx

          λ = 8 cm

6 0
3 years ago
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie
Nutka1998 [239]

Answer:

i = 0.5 A

Explanation:

As we know that magnetic flux is given as

\phi = NBA

here we know that

N = number of turns

B = magnetic field

A = area of the loop

now we know that rate of change in magnetic flux will induce EMF in the coil

so we have

EMF = NA\frac{dB}{dt}

now plug in all values to find induced EMF

EMF = (20)(50 \times 10^{-4})(\frac{6 - 2}{2})

EMF = 0.2 volts

now by ohm's law we have

current = \frac{EMF}{Resistance}

i = \frac{0.2}{0.40} = 0.5 A

5 0
3 years ago
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