Subtract the following and give the answer using significant figures 3.400-0.7591

Unpolarized light passes through three polarizing filters. The first one is oriented with a horizontal transmission axis, the se

Xelga [282]

Answer:

10.028%

Explanation:

$\theta$ = Angle between polarizer

The polarized light after passing through first polarizer

$I=\frac{I_0}{2}$

The polarized light after passing through second polarizer

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow I_2=\frac{I_0}{2}cos^231.8\\\Rightarrow I_2=0.36115I_0$

The polarized light after passing through third polarizer

$I_3=I_2cos^2(90-31.8)\\\Rightarrow I_3=0.36115I_0cos^2(90-31.8)\\\Rightarrow I_3=0.10028I_0$

$\frac{I_3}{I_0}\times 100=\frac{0.10028I_0}{I_0}\times 100\\ =10.028\ \%$

The percent of the light gets through this combination of filters is 10.028%

4 0

3 years ago

This is one of the four planets Mercury,Venus,Earth,and Mars, whose orbits lie nearest the Sun

galben [10]

Answer:

Mercury

Explanation:

Mercury is the closest planet to the sun followed be Venus. Though Mercury is the closest Venus is actually the hotest planet

6 0

3 years ago

Read 2 more answers

A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i

Lesechka [4]

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

7 0

3 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

3 years ago

Please someone help me !!

sergij07 [2.7K]

Answer:

D

Explanation:

because if the solvent is more than the solvent then we can't resolve it.

so our product will be suspended

6 0

3 years ago