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3 years ago

Does aluminum foil refract light? Why?

Physics

1 answer:

nikitadnepr [17]3 years ago

4 0

yes it does it because the material cause it to refract and the stronger it is the more it will refract

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Which of the following is true of microscopic particles? They can be localized or not localized in space They can explain the ph

Alona [7]

Answer:

true about microscopic particle is that they can explain the phenomena of diffraction and interference.

Explanation

the first thing mentioned in question particles are localized in space according to classical physics particle can be well localized in space because its momentum and position can be determine simultaneously .second their properties can be measured and they can act like wave particle by particle build up a wave. so all of they things mention above only thing true about particles is that they describe the phenomena of diffraction and interference.

5 0

3 years ago

What is most likely to happen during deposition?

Iteru [2.4K]

Answer:

a person appears at a specified time and place and gives sworn testimony

Explanation:

Hope this helps

6 0

3 years ago

calculate the amount of work done when a grocery store stocker uses 120N of force to lift a sack of flour 1.5m onto a shelf

marta [7]

Hello There!

Answer is provided in the image attached.

Have a great day!

4 0

3 years ago

A 0.500 m length of wire with a cross-sectional area of 3.14 * 10 ^ - 6 meters squared is found to have a resistance of 2.53 * 1

mr_godi [17]

Answer:

The wire is made of silver (ρ = 1.59×10⁻⁸ ohms/m)

Explanation:

Applying,

R = ρL/A................. Equation 1

Where R = Resistance length of the wire, ρ = Resistivity of the wire, L = Length of the wire, A = crosssectional area of the wire

make ρ the subject of the equation

ρ = RA/L............. Equation 2

From the question,

Given: R = 2.53×10⁻³ ohms, A = 3.14×10⁻⁶ m², L = 0.5 m

Substitute these values into equation 2

ρ = (2.53×10⁻³)(3.14×10⁻⁶)/0.5

ρ = 1.59×10⁻⁸ ohms/m

Hence from the resistivity chart, the wire is made of silver

7 0

3 years ago

In Fig. P24.59, each capacitance C1 is 6.9 mF, and each capacitance C2 is 4.6 mF. (a) Compute the equivalent capacitance of the

mrs_skeptik [129]

(a) Equivalent capacitance of network between points a and b is 2.3μF.

(b) Charge on each of the three capacitors nearest a and b is 920 μC.

A) Let's consider the right side that is farthest from the point 'a'.

Three C1 are connected in series

C= 1/C1 + 1/C1 +1/C1 = C1/3

C = 6.9/3 = 2.3μF

Now this C is connected in parallel with C2

So, C= 2.3 + 4.6 = 6.9μF

Again we get three capacitors of 6.9μF each, connected in series.

C = C1/3 = 2.3μF

It again combines with 4.6μF in parallel

C = 4.6 + 2.3 =6.9μF

Now, this final reduction is the same as that of the first.

In the end, we have 3 capacitors of 6.9μF each connected in series.

Equivalent Capacitance C(eq) = C1/3 = 6.9/3 = 2.3μF

= 2.3 × $10^{-6}$ F

B) C(eq) = 2.3 × $10^{-6}$ F

Vab = 400 V (Given)

Charge on each of the three capacitors nearest a and b (Q) =?

Q = C(eq) × Vab

= 2.3 × $10^{-6}$ × 400 = 9.2 × $10^{-4}$ C = 920μC

Hence, the equivalent capacitance of the network between points a and b is 2.3μF.

And Charge on each of the three capacitors nearest a and b is 920 μC.

Learn more about Capacitance here brainly.com/question/13578522

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8 0

2 years ago