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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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Answer:

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

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using equation of specific heat  

Q = m S ΔT

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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