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Oxana [17]
2 years ago
5

Inside a material with an index of refraction not equal to 1, both the speed and the wavelength of light must change. is the wav

elength of light shorter or longer in water then in air?
Physics
1 answer:
9966 [12]2 years ago
4 0

Inside a material with an index of refraction not equal to 1, both the speed and the wavelength of light must change. The wavelength of light is shorter in water than in air.

The light's frequency doesn't change. The incident light deviates due to variations in light's speed and wavelength. When light moves from a rare to a denser medium, its speed and wavelength are reduced.

Speed of light is given by ;

λf = c

where, λ = wavelength

            f = frequency

            c = speed of light

f = c / λ

λ = c / f

Since energy doesn't change when light travels from air to water, so does its frequency. However, when light travels faster, its wavelength gets shorter. Thus, the wavelength of light is less in water than in air even though its frequency is the same when it travels from air to water.

Learn more about speed of light here;

brainly.com/question/1555553

#SPJ4

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Answer:

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Explanation:

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The momentum of an object is determined by the object’s mass and velocity. Which object has more momentum?
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PLEASE HELP! URGENT<br><br> Explain how the forces need to change so an aeroplane can land
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Relative humidity _______.
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At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q
Bad White [126]

Answer:

\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}

\displaystyle \theta =68^o

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

\displaystyle F=\frac{KQ_1Q_2}{d^2}

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}

\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l

The force F3 is

\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle

F2 is pointed upwards and its horizontal component is zero

\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle

\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

Now we compute the total force

\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}

\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle

\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}

Now we compute the magnitude

\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}

The direction of the total force is given by

\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558

\boxed{\displaystyle \theta =68^o}

6 0
3 years ago
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