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3 years ago

How can two polarizing filters be used to show that light has some properties of a wave

1 answer:

6 0

Polarizing filters: used to show light has some properties of a wave in that way of property is that light can be thought of as traveling forward in waves, with the <span>wave. sorry if this is too late but google is good</span>

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the specific heat of gold is 0.031 calories degrees Celsius and the specific heat of silver is 0.057 calories degrees Celsius so

givi [52]

Answer:

Explanation:

Given:

Specific heat of gold = 0.031cal/°C

Specific heat of silver = 0.057cal/°C

To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.

It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.

Now,

The higher the specific heat capacity the more energy it is required to heat up the substance.

So, Gold with a specific heat capacity of 0.031cal/°C will heat up faster.

8 0

2 years ago

S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

3 years ago

There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

5 0

3 years ago

A man can jog 10 miles in 90 minutes. What’s his speed in mph?

Yuki888 [10]

Answer:

hi there!

the correct answer to this question is: 6.67 mph

Explanation:

you convert minutes to hours

10 miles * 60 mins / 90 mins

7 0

2 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago