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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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An object increases vert its velocity from 22M/S 236M/S and five seconds. What is the acceleration of the object

MariettaO [177]

Answer:

$\boxed {\boxed {\sf a=42.8 \ m/s^2}}$

Explanation:

Acceleration can be found by dividing the change in velocity by the time.

$a=\frac{v-u}{t}$ (v is the final velocity, u is the initial velocity, t is the time).

The velocity increased from 22 m/s to 236 m/s in 5 seconds. Therefore:

$v=236 \ m/s\\u=22 \ m/s\\t= 5 \ s$

Substitute the values into the formula.

$a=\frac{236 \ m/s - 22 \ m/s}{5 \ s}$

Subtract in the numerator.

236 m/s-22 m/s=214 m/s

$a=\frac{214 \ m/s}{5 \ s}$

Divide.

$a=42.8 \ m/s^2$

The acceleration of the object is 42.8 meters per square second.

6 0

3 years ago

It takes 399 newtons to move a 7.2 kg object. What is the acceleration of the object?

Nookie1986 [14]

Answer:

a = 55.42m/s²

Explanation:

F = m*a

F = force(Newtons)

m = mass

a = acceleration

a = F/m

a = 399N/7.2kg

a = 55.42m/s²

8 0

2 years ago

Which pair below describes isotopes of the same element? A) an atom with 6 protons and 8 neutrons - an atom with 8 protons and 6

iogann1982 [59]

An isotope is an atom with a different number of neutrons than another atom of the same element. Since atoms of the same element all have the same number of protons, choice B(6pro &6neu vs. 6pro&7neu) is an example of isotopes

7 0

3 years ago

Read 2 more answers

What is the answer to this question

leva [86]

The answer is A 0 degrees is the freezing point of water

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7 0

3 years ago

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago