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Aleksandr-060686 [28]
2 years ago
11

Types of polarization physics

Physics
1 answer:
Sidana [21]2 years ago
8 0

Answer:

Linear polarization.

Circular polarization.

Elliptical polarization.

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After a wave passes through a medium, particles in the medium
sertanlavr [38]

Answer:

If a particle is affected by a wave, then the particles are displaced. They move along the direction of the wave. Hence, After a wave passes through a medium, particles in the medium are moved along with the wave.

Explanation:

hope this helps you

3 0
3 years ago
What does the addition of two vectors give u
sweet-ann [11.9K]
Addition of 2vector gives you 1large vector quantity
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A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

negative indicates distance is decreasing with time

5 0
3 years ago
The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end
Darina [25.2K]

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

T = mg

here we will have

T = 2(9.8) = 19.6 N

now we will have speed of wave given as

v = \sqrt{\frac{T}{\mu}}

here we will have

v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}

v = 16.33 m/s

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

\lambda = \frac{v}{F}

\lambda = \frac{16.33}{100} = 0.16 m

so wavelength will be 0.16 m

5 0
3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

\mathrm{m}_{2}=900 \mathrm{kg}

\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

600 \times 5=900 \times v_{2}

3000=900 \times v_{2}

\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
4 years ago
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