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EastWind [94]
3 years ago
10

Which statement is true about an airplane wing during flight?

Physics
2 answers:
azamat3 years ago
5 0

c. air above the wing travels faster than air below the wing.

allochka39001 [22]3 years ago
3 0
C. air above the wing travels faster than air below the wing.
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CO2 and N2O keep the energy that gets to Earth from the sun inside the atmosphere. Without greenhouse gases, our planet would be too cold. But due to the recent increase in greenhouse gases, more energy released from the sun is contained in the Earth, heating it up.
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What transfers energy in the form of vibrating electric and magnetic fields?
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Electromagnetic waves are waves that consist of vibrating electric and magnetic fields. They transfer energy through matter or across space. The transfer of energy by electromagnetic waves is called electromagnetic radiation. ... The two vibrating fields together form an electromagnetic wave.
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3 years ago
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The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were
Simora [160]

Answer:

a) 45571 N  

b) 22786 N

c) 4557 N

Explanation:

  • Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
  • In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

       \Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)

  • Now, just applying the original form of  Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

       \Delta p = F* \Delta t  (2)

  • So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

       F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)

       F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)

       F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)

4 0
3 years ago
An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1
Ksivusya [100]
<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

                 F=\frac{GMm}{r^2}

           Where G =  6.67 x 10⁻¹¹ N m²/kg²

                       M = Mass of body 1

                       M = Mass of body 2

                       r = Distance between them

Here we have

                 M = Mass of Sun = 1.99×10³⁰ kg

                 m = Mass of asteroid = 4.00×10¹⁶ kg

                 F = 3.14×10¹³ N

Substituting

                   F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0
3 years ago
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