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3 years ago

Jalen combines two substances in science class. Once combined, only one of the two substances

Physics

1 answer:

dusya [7]3 years ago

4 0

Answer: b. The combination is a mixture because the substances can be separated

Explanation: Based on the facts presented above, the combination of both both substances can be referred to as a mixture due to the following:

A mixture is obtained when two or more substances or materials are combined without a chemical reaction. This is observed when Jalen combined substance 1 and 2 with only one of the substances becoming visible after the combination.

The other reason is that, a mixture can be separated back into its original constituent, this is evident when the combination was filtered with only substance 2 going through the filter and substance 1 remaining in the filter

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Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

$U=k\frac{q_1q_2}{r_{1,2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

$U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}$ (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

$U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J$

The electric potential energy between the three charges is 80.91 J

7 0

3 years ago

Someone help me, I'm stuck

Bas_tet [7]

C is the answer hope that helps you

3 0

3 years ago

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

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3 years ago

A piece of silicon sample has a resistivity of 0.1 ω.Cm. Its thickness is 100µm. The electron mobility is 1350cm 2 v -1sec-1. Wh

lawyer [7]

The answer is Rh = 135 cm^3 and B = 0.05185 wh/m^2

Explanation:

Resitivity of silicon = 0.1

thickness = 100um

so, I = ma

Required to find out concentration of electron , we know that

Rh = up

By putting in the values,

Rh = 1350 x 0.1

Rh = 135 cm^3

Now consider,

Rh = 1 / Rh.q

= 1 / Rh . q

= 1 / 135 x1.609 x10^-19

= 4.6037 x 10^16 / cm^3

Vh = BIRh / w

B = Vh w/ IRh

B = -70 x10^-6 x 100 x10^-6 / 1x 10^-3 x 135 x 10^-6

B = 0.05185 wh / m^2

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3 years ago

Some students fire a rocket of mass 0.150kg into the air. If at the top of it's flight it has a gravitational potential energy o

mamaluj [8]

PE=mgh
150J=(0.150)*(9.8)*h
h=102m

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3 years ago