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3 years ago

Which color of light is diffracted at a greater angle from a diffraction grating, red or yellow light?

1 answer:

3 0

The amount of diffraction depends on the wavelength of light, with shorter wavelengths being diffracted at a greater angle than longer ones (in effect, blue and violet<span> light are diffracted at a larger angle than is red light).

I hope my answer has come to your help. God bless and have a nice day ahead!
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A sinusoidal wave travels along a string. if the time for a particular point to move from maximum displacement to zero displacem

Lisa [10]

The time it takes for a point to reach 0 displacement from maximum displacement is $\frac{\pi}{2}$ of a cycle. Thus, one period ( $2\pi$ ) will take $T=0.17*4=0.68s$ .

The frequency is simply $f=\frac{1}{T}=\frac{1}{0.68}=1.47Hz$ .

The speed is given by: $v=f\lambda=1.47*1.4 = 2.06m/s$

7 0

3 years ago

An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra

WARRIOR [948]

Answer:

The inventors claim is not real

a) No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

The power input is $P_i = 0.25 kW = 0.25 *10^{3} \ W$

The rate of heat transfer $J = 3050 J/s$

The temperature of the freezer content is $T = 270 \ K$

The ambient temperature is $T_a = 293 \ K$

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

$COP = \frac{T }{Ta - T}$

substituting values

$COP = \frac{270 }{293 - 270}$

$COP =11.7$

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

$COP = \frac{J}{P_i}$

substituting values

$COP = \frac{3050}{0.25 *10^{3}}$

$COP = 12.2$

Now given that the COP of an ideal refrigerator is less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

4 0

3 years ago

(TCO-8) A band-pass filter has fC1 = 5 kHz and fC2 = 88 kHz. Calculate Bandwidth (BW) and center frequency (fo) for this filter.

Temka [501]

Bandwidth is the difference between the upper and lower frequencies in a continuous band of frequencies.

Mathematically can be expressed as,

$BW = F_{c2}-F_{c1}$

The upper frequency is 88Hz and the lower frequency is 6, therefore the Bandwidth would be,

$BW = (88-5)kHz$

$BW = 83kHz$

The center frequency is given on the basis of the square root of the multiplication of the two reference frequencies, then,

$f_0 = \sqrt{f_{c1}*f_{c2}}$

$f_0 = \sqrt{88*5}$

$f_0 = 20.97kHz$

7 0

3 years ago

a block measuring 20cm by 10cm by 5cm rests on a flat surface. The block has a weight of 3N. Determine the maximum pressure it e

Lady_Fox [76]

Max preassure = force / min area
= 3N / 0.1 x 0.05
= 600N/m(squared)
Copy off of the picture below itll help better, its what someone sent me when i asked this question