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3 years ago

What is the difference between polarized light and unpolarized light?

1 answer:

5 0

Unpolarized light has light waves that are oriented in all directions while polarized light has all the light waves oscillating in one direction. An example for unpolarized is the Sun, or any other simple type of light.

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When two positive charges are brought close together, what happens to the

telo118 [61]

Answer:

they will move away from each other

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3 years ago

An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?

snow_lady [41]

The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.

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2 years ago

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

6 0

3 years ago

The measured value of the speed of light is about 300,000 km/s. suppose a futuristic space train is traveling at 200,000 km/s wi

Gwar [14]

The speed of light generally would be 300000km/s but since the train is moving in the same direction as the light it would apparently appear to be 100000km/s

3 0

3 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

4 years ago