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KIM [24]
3 years ago
7

A substance has a volume of 4,000 cm and a mass that is 2,000 g. What is the density of the substance.

Physics
1 answer:
podryga [215]3 years ago
5 0
Density is mass divided by volume making the answer 0.5 g/cm^3
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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5
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Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E =  K(q/d²)

where K = 8.99e9

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Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

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E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

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= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

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