As an airplane moves through the air, its wings cause changes in the speed and pressure of the air moving past them. These changes result in the upward force called lift.

The Bernoulli principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in the pressure exerted by the fluid.

A wing is shaped and tilted so the air moving over it moves faster than the air moving under it. As air speeds up, its pressure goes down. So the faster-moving air above exerts less pressure on the wing than the slower-moving air below. The result is an upward push on the wing—lift!

4 0

3 years ago

A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction

Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

Index refraction = 1.47

(a). We need to calculate the speed

Using formula of speed

$n = \dfrac{c}{v}$

Where, n = refraction index

c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

$1.47=\dfrac{3\times10^{8}}{v}$

$v=\dfrac{3\times10^{8}}{1.47}$

$v= 2.04\times10^{8}\ m/s$

(b). We need to calculate the wavelength

Using formula of wavelength

$n=\dfrac{\lambda_{0}}{\lambda}$

$\lambda=\dfrac{\lambda_{0}}{n}$

Where, $\lambda_{0}$ = wavelength in vacuum

$\lambda$ = wavelength in medium

Put the value into the formula

$\lambda=\dfrac{650\times10^{-9}}{1.47}$

$\lambda=442\times10^{-9}\ m$

Hence, The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

3 0

3 years ago

As the distance between two charged objects increases, the strength of the electrical force between the objects

AveGali [126]

As the distance between two charged objects increases, the strength of the electrical force between the objects <em>decreases</em>.

7 0

3 years ago

Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri

Artemon [7]

First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

Ep=m*g*h=0.7*9.8*7=48.02 J

The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick

4 0

3 years ago