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4 years ago

15

For what visible wavelengths of light do the reflected waves interfere constructively? The range of wavelength of visible light

is from 380 nm to 750 nm.

1 answer:

Otrada [13]4 years ago

7 0

Answer:

None of the wavelength is in the visible range

Explanation:

Constructive interference of the reflected waves for different wavelengths can be estimated using:

λ $_{m}$ = 2nd/m

where m is 1,2,3, ...

Therefore:

m=1, λ $_{1}$ = 750 nm

m=2, λ $_{1}$ = 750/2 = 375 nm

The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.

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What is the difference between observation and hypothesis

jekas [21]

Answer:

Observation- The action or process of observing something or someone carefully or in order to gain information.

Hypothesis-a proposed explanation made on the basis of limited evidence as a starting point for further investigation.

Explanation:

8 0

3 years ago

A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a

marshall27 [118]

Explanation:

The given data is as follows.

mass (m) = 0.160 kg, spring constant (k) = 8 n/m,

Maximum speed ( $v_{m}$ ) = 0.350 m/s

Formula for angular frequency is as follows.

$\omega = \sqrt{\frac{{k}{m}}$

$\omega = \sqrt{\frac{{8}{0.160}}$

$\omega$ = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

$\nu_{max} = A \omega$

A = $\frac{\nu}{\omega}$

= $\frac{0.35}{7.07}$

= 0.05 m

Hence, value of amplitude is 0.05 m.

(b) Displacement = 0.030 m

Formula for mechanical energy is as follows.

M.E = $\frac{1}{2}kA^{2}$

Putting the values into the above formula as follows.

M.E = $\frac{1}{2}kA^{2}$

= $\frac{1}{2} \times 8 \times (0.05)^{2}$

= $9.8 \times 10^{-3}$ Joule

For x = 0.03,

As, P.E = $\frac{1}{2} \times kx^{2}$

= $\frac{1}{2} \times 8 \times (0.03)$

= $3.6 \times 10^{-3}$

Hence, calculate the kinetic energy as follows.

K.E = M.E - P.E

= ( $9.8 \times 10^{-3}$ - $3.6 \times 10^{-3}$ ) J

= $6.2 \times 10^{-3}$ J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is $6.2 \times 10^{-3}$ J.

7 0

3 years ago

if you apply a Force of F1 to area A1 on one side of a hydraulic jack, and the second side of the jack has an area that is twice

Firlakuza [10]

Answer:

$2F_{1}$

Explanation:

F₁ = Force on one side of the jack

A₁ = Area of cross-section of one side of the jack

F₂ = Force on second side of the jack

A₂ = Area of cross-section of second side of the jack = 2 A₁

Using pascal's law

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{A_{_{2}}}$

$\frac{F_{1}}{A_{1}}= \frac{F_{_{2}}}{2A_{_{1}}}$

$F_{1}= \frac{F_{2}}{2}\\$

$F_{2}= 2F_{1}$

7 0

3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

bija089 [108]

Answer:

$v_o\approx0.7059\ m.s^{-1}$

Explanation:

Given:

mass of the boat, $m=120\ kg$

uniform speed of the boat, $v=1\ m.s^{-1}$

rate of accumulation of water mass in the boat, $\dot m=100\ kg.hr^{-1}$

time of observation, $t=0.5\ hr$

The mass of the boat after the observed time:

$m_o=m+\dot m\times t$

$m_o=120+100\times 0.5$

$m_o=170\ kg$

<u>Now using the conservation of momentum:</u>

$m.v=m_o.v_o$

$120\times 1=170\times v_o$

$v_o\approx0.7059\ m.s^{-1}$

4 0

3 years ago

The density of blood is 1.05 kg/m3, find the mass of a bag of blood for a transfusion, if the volume is 1.5 m3.

hoa [83]

(1.5 m^3) • (1.05 kg/m^3) = 1.575 kg. That's quite a bag you've got there ! 1 m^3 is like 264 gallons of blood. Hope the poor patient survives the transfusion. Also, the actual density of blood is around 1.05 g/cm^3, or 1050 kg/m^3. The blood they're giving the guy in this question is about 18% less dense than the AIR in his hospital room, and they're pumping 264 gallons of it into him. Maybe THAT'S his whole problem.

6 0

3 years ago

Read 2 more answers