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3 years ago

The index of refraction of a material medium must be less than 1 true or flase

Physics

1 answer:

OLga [1]3 years ago

8 0

That statement is flase. The index of refraction
of every material is more than 1.0 .

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A vector always consits of

Kazeer [188]

Answer:

Size and Direction

Explanation:

8 0

3 years ago

If you traveled one mile at a speed of 100 miles per hour and another mile at a speed if 1 mile per hour, your average speed wou

Semenov [28]

Answer:

v = 1.98 mph

Explanation:

Given that,

Speed to travel one mile is 100 mph

Speed to travel another mile is 1 mph

The formula used to find your average speed is given by :

$v=\dfrac{2v_1v_2}{v_1+v_2}$

Putting the values, we get :

$v=\dfrac{2\times 100\times 1}{100+1}$

v = 1.98 mph

So, yours average speed is 1.98 mph.

7 0

3 years ago

This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

$P = \frac{1}{2} \rho v^3 A C_p$

where $\rho = 1.2 kg/m^3$ is the air density, v = 8.89 m/s is the wind speed, A is the swept area and $C_p = 0.45$ is the efficiency

$10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45$

$10000 = 190A$

$A = 10000 / 190 = 52.7 m^2$

The swept area is a circle with radius r being the blade length

$\pi r^2 = A = 52.7$

$r^2 = 52.7 / \pi = 16.79$

$r = \sqrt{16.79} = 4.1 m$

4 0

3 years ago

this refers to the process of manufacturing that introduced powered machinery to the production of goods

Sidana [21]

This had to do with gain power and trade inequality business

5 0

3 years ago

One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit

nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

$\lambda$ = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

$\theta=\dfrac{1.22\lambda}{D}$

The width is given by

$d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m$

The required width is 0.000003782 m

Minimum resolvable line separation is given by

$\dfrac{0.000003782}{2}=0.000001891\ m$

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when $\lambda=157\ nm$

$d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m$

The new minimum resolvable line separation between adjacent lines is

$\dfrac{0.00000239425}{2}=0.000001197125\ m$

6 0

3 years ago