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3 years ago

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You have 50L of water, it froze. So find its new volume. Density of water is 1000kg/m^3 or 1kg/L, density of ice is 920kg/m^3 or

0.92kg/L.

1 answer:

ad-work [718]3 years ago

4 0

Answer:

D (density) = Mass / Volume

V (ice) = Mass / Density = 50000 g / .92 kg / L

V = 50 kg / .92 kg / L

1 Liter of water weighs 1 kilogram = 50 L weighs 50000 g

V = 54.3 L the mass does not change upon the change of phase (freezing)

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

b)

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

c)

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

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