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4 years ago

12

What happens to a light beam that passes through a polarizing crystal? What happens when plane-polarized light passes through a

second polarizing crystal set perpendicular to the first crystal?

1 answer:

levacccp [35]4 years ago

5 0

Answer:

it gives rise to vibrating in one plane.when light passes through second polarizing crystal no light passing through.

Explanation:

in the second polarizing crystal only darkness emerges.

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A book with mass 2.3 kg sits on a table. What is the normal force on the

natita [175]

Explanation:

The table is level and there are no other forces on the book, so the normal force is equal to the weight.

N = mg

N = (2.3 kg) (9.8 m/s²)

N = 22.5 N

6 0

3 years ago

Read 2 more answers

A star produces 2x10^26 watts. how much energy does it lose every minutes

Step2247 [10]

Answer:

Energy loss per minute will be $120\times 10^{26}j$

Explanation:

We have given the star produces power of $2\times 10^{26}W$

We know that 1 W = 1 J/sec

So $2\times 10^{26}W=2\times 10^{26}J/sec$

Given time = 1 minute = 60 sec

So the energy loss per minute $=2\times 10^{26}\times 60=120\times 10^{26}j$

We multiply with 60 we have to calculate energy loss per minute

7 0

4 years ago

She left the cubes in the water for three hours which of the following describes a heat flow that took place during those three

quester [9]

Answer:

veffvevfevve

Explanation:

3 0

3 years ago

Read 2 more answers

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

What is the difference between a series circuit and a parallel circuit?

Angelina_Jolie [31]

Parallel has more than one circuit or form of energy

series has only one form of energy circuit

6 0

3 years ago