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3 years ago

15

The wavelength of a light wave is 700 nm in air; this light appears red. If this wave enters a pool of water, its wavelength bec

omes λair/n = 530 nm. If you were swimming underwater, the light would still appear red. Given this, what property of a wave determines its color?

1 answer:

hodyreva [135]3 years ago

6 0

Answer

given,

wavelength of light in air = 700 nm

Wavelength of light in water = 530 nm

We know that speed of light changes when it moves from one medium to another.

And the frequency of the wavelength does not changes if the medium changes.

we also know that,

v = ν λ

From the above equation we can say that if frequency is constant so, with the change in velocity changes wavelength will also change.

Hence, wavelength is the property of the wave which determines color.

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Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti

Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

$\dfrac{dF_{net}}{df}=0$

We need to calculate the unknown force

Using formula of net force

$\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}$

Put the value into the formula

$\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}$

$\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}$

The magnitude of net force,

$F_{net}=\sqrt{F_{x}^2+F_{y}^2}$

$F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}$

$F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}$

$F_{net}=\sqrt{F^2+4899.78+36.232F}$

On differentiating w.r.to F

$(\dfrac{dF_{net}}{dF})^2=2F+36.232$

$0=2F+36.232$

$F=-\dfrac{36.232}{2}$

$F=-18.116\ lb$

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

6 0

2 years ago

A 60.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 20.0 m. (a) What is the

IrinaVladis [17]

Answer:

3 because

Explanation:

60÷20=3 ans is this

3 0

2 years ago

6) If a mass of an object is decreased to half and acting force is reduced by quarter the acceleration of its motion

Zepler [3.9K]

Answer:

Decreases to half.

Explanation:

From the question given above, the following data were obtained:

Initial mass (m₁) = m

Initial force (F₁) = F

Initial acceleration (a₁) =?

Final mass (m₂) = ½m

Final force (F₂) = ¼F

Final acceleration (a₂) =?

Next, we shall determine a₁. This can be obtained as follow:

F₁ = m₁a₁

F = ma₁

Divide both side by m

a₁ = F / m

Next, we shall determine a₂.

F₂ = m₂a₂

¼F = ½ma₂

2F = 4ma₂

Divide both side by 4m

a₂ = 2F / 4m

a₂ = F / 2m

Finally, we shall determine the ratio of a₂ to a₁. This can be obtained as follow:

a₁ = F / m

a₂ = F / 2m

a₂ : a₁ = a₂ / a₁

a₂ / a₁ = F/2m ÷ F/m

a₂ / a₁ = F/2m × m/F

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

From the illustrations made above, the acceleration of the car will decrease to half the original acceleration

7 0

3 years ago

PLSSSSS HELP!!!!!!! ASAP!!!!! MARK BRAINLIEST!!!!

Ymorist [56]

Answer: I would say the answer is B.

Explanation: It looks and sounds the most correct.

5 0

2 years ago

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

sesenic [268]

Work done by a given force is given by

$W = F.d$

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = $W_1 = Fdcos\theta$

$W_1 = 12*5cos15$

$W_1 = 57.96 J$

Now similarly work done by frictional force

$W_2 = Fdcos\theta$

$W_2 = 4*5cos180$

$W_2= -20 J$

Now total work done on sled

$W_{net}= W_1 + W2$

$W_{net} = 57.96 - 20 = 37.96 J$

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3 years ago