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3 years ago

What is the mass of an object with a density if 4.0g/cm3 that displaces 3.0cm3 of water?

1 answer:

7 0

I am assuming you mean the density is 4.0 g/ cm3 and the displaced H20 is 3.0 cm3 (3 mls when converted to cm3). Look at the units that should tell you how to solve the problem. Your answer must in g (mass), thus you take 4.0 g/cm3 times 3.0 cm3. Units are treated like numbers, for example 3 times 1/3 equals 1 - both 3's cancel leaving only 1 time 1 which equals 1. In this case you have 4 g/cm3 times 3.0 cm3, cm3's cancel leaving only g as your unit at the end. The rest is for you to solve! Enjoy your problem. I am assuming you mean the density is 4.0 g/ cm3 and the displaced H20 is 3.0 cm3 (3 mls when converted to cm3). Look at the units that should tell you how to solve the problem. Your answer must in g (mass), thus you take 4.0 g/cm3 times 3.0 cm3. Units are treated like numbers, for example 3 times 1/3 equals 1 - both 3's cancel leaving only 1 time 1 which equals 1. In this case you have 4 g/cm3 times 3.0 cm3, cm3's cancel leaving only g as your unit at the end. The rest is for you to solve! Enjoy your problem.

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3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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3 years ago