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2 years ago

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Solve this worksheet. Need this worksheet with Answer

2 answers:

Ray Of Light [21]2 years ago

8 0

There’s nothing, it’s blank

andrey2020 [161]2 years ago

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There’s nothing there? it’s all blank..

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Why does refracting light bend when it enters a denser or less dense medium?

Rzqust [24]

Answer: The light bends because light travels fast but it slows down in a denser medium. For example light refracts in water or it bends after passing through air. When light passes through air ( a less dense medium ) then through water ( a more dense medium ) the beam of light bends because light travels more slowly in a denser medium then it picks up its pace again once it passes. The density of the substance determines how much the light is refracted. I hope this makes sense and I hope this answered your question!! :)

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3 years ago

Rafe is placing his car on a lift to change the oil and oil filter. He needs to drive up the lift 1.5 meters and the lift raises

jeyben [28]

Answer:

104 IMA

Explanation:

7 0

2 years ago

he product side of a chemical reaction is shown. → 7Ti2(SO4)3 Which number represents a coefficient? 2 3 4 7

valkas [14]

I think it would be 7

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3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

or

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago

If 3.61 m3 of a gas initially at STP is placed under a pressure of 2.67 atm , the temperature of the gas rises to 37.9 ∘C. What

Pavel [41]

Answer: $1.54m^3$

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas at STP = 1 atm

$P_2$ = final pressure of gas = 2.67 atm

$V_1$ = initial volume of gas = $3.61m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas at STP = $0^oC=273+0=273K$

$T_2$ = final temperature of gas = $37.9^oC=273+37.9=310.9K$

Now put all the given values in the above equation, we get:

$\frac{1atm\times 3.61m^3}{273K}=\frac{2.67\times V_2}{310.9K}$

$V_2=1.54m^3$

Thus the final volume will be $1.54m^3$

7 0

3 years ago