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4 years ago

Is a process that modifies light waves so they vibrate in a single plane

Physics

1 answer:

madam [21]4 years ago

8 0

The process you're fishing for is "polarization", but that's a

misleading description.

Polarization doesn't do anything to change the light waves.

It simply filters out (absorbs, as with a polarizing filter) the

light waves that aren't vibrating in the desired plane, and

allows only those that are to pass.

The intensity of a light beam is always reduced after

polarizing it, because much (most) of the original light

has been removed.

A laser light source may be thought of as an exception,

since everything coming out of the laser is polarized.

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On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question

romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs $\frac{100}{0.37}=270.27kg$

8 0

4 years ago

Bob, Jill, Kim, and Steve measure an object's length, density, mass, and brightness, respectively. Which student must derive a u

netineya [11]

The answer is A. Bob (object's length)

3 0

3 years ago

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

sashaice [31]

Answer: The average atomic mass of the given element is 20.169 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

$\text{Average atomic mass}=20.169amu$

Hence, the average atomic mass of the given element is 20.169 amu.

4 0

3 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

3 years ago

System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas

Anna [14]

Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

$F=G \frac{m_1 m_2}{r^2}$

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

$F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}$

system B has masses m and 2m separated by a distance 2r:

$F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}$

system C has masses 2m and 3m separated by a distance 2r:

$F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}$

system D has masses 4m and 5m separated by a distance 3r:

$F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}$

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

5 0

3 years ago