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2 years ago

Circuit has a single pathway for the electrons of flow. series parallel

Physics

2 answers:

Bad White [126]2 years ago

5 0

Series circuit!!!!!!!!!!

Alex787 [66]2 years ago

3 0

Series circuit has so a single path way for the electrons to flow

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Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through

Norma-Jean [14]

Answer:

1. $\theta=29.84^{0}$

2. $\theta=60.15^{0}$

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of first polarizer

which is given as

$I=\frac{I_{0} }{2}$

$I= \frac{655\frac{W}{M^{2} } }{2}$

$I=327.5\frac{W}{m^{2} }$

For a smaller angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} Cos^{2}(90^{0} - \theta)$

$I_{2} =I_{1} sin^{2}\theta$

$\frac{I_{2} }{I_{1} }=Sin^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = sin\theta$

$\theta=Sin^{-1}(0.4977)$

$\theta=29.84^{0}$

For a larger angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} cos^{2}\theta$

$\frac{I_{2} }{I_{1} }=Cos^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = cos\theta$

$\theta=Cos^{-1}(0.4977)$

$\theta=60.15^{0}$

7 0

2 years ago

How do you find the change in potential energy

ratelena [41]

P.E = mgh

This is the formula for potential energy.

This is where m is mass, g is the acceleration due to gravity, and h is height.

All you have to do is multiply all these numbers together.

3 0

2 years ago

Please help fast i need urgent

Sloan [31]

Is there more information ?

5 0

2 years ago

Anyone know which wire matches the other one? for all 4. Need help! Thanks :-)

Fittoniya [83]

red goes to red, black goes to white, yellow goes to green, blue goes to blue.

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3 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

2 years ago