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3 years ago

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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In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocit

ivolga24 [154]

Answer:

$m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg$

Explanation:

The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed

$m_{tackler}*v_{tackler}=(m_{tackler}+m_{receiver})V$ (1)

We solve in order to find the receiver mass:

$m_{receiver}={m_{tackler}*v_{tackler}/V}-m_{tackler}$

$m_{receiver}=115Kg*3.1/(1.6)-115Kg=107.8Kg$

5 0

3 years ago

Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H

djyliett [7]

Answer:

The wavelength of the infrared wave is <u>0.0001 m</u>.

Explanation:

Given:

Frequency of an infrared wave is, $f=3.0\times 10^{12}\ Hz$

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

$v=c=3.0\times 10^8\ m/s$

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

$v=f\lambda$

Where, $\lambda \to \textrm{Wavelength of infrared wave}$

Now, rewriting the above formula in terms of wavelength, $\lambda$ , we get:

$\lambda=\dfrac{v}{f}$

Now, plug in $3.0\times 10^8$ for 'v', $3.0\times 10^{12}$ for 'f' and solve for $\lambda$ . This gives,

$\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m$

Therefore, the wavelength of the infrared wave is 0.0001 m.

5 0

3 years ago

Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called

Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.

So, these types of waves are called longitudinal waves.

2. d. a medium

There are two types of waves:

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium

3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0

3 years ago

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$