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padilas [110]
2 years ago
15

Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 2

0 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume?
Physics
2 answers:
alexandr402 [8]2 years ago
8 0

Answer:

5.08 but if it asks that it needs to be rounded to the nearest whole number, then the answer would be 5

Explanation:

I got this wrong because I was supposed to round it :( Don't my mistake guys

lbvjy [14]2 years ago
6 0
Density = mass/volume, volume = mass/density.
Since the mass of the small cube equals 20 and the mass of the large cube is double it would be 40.
Now plug in volume = 40 g/(7.87 g/cm^3).
Thus giving you a volume 5.08 cm^3
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A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

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\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

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\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

Thus from eqn (2), the total charge is given by:

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

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