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GalinKa [24]
3 years ago
8

The side length of a 243-gram copper cube is 3 centimeters. Use this information to write a model for the radius of a copper sph

ere as a function of its mass. Then, find the radius of a copper sphere with a mass of 50 grams. How would changing the material affect the function?
Physics
1 answer:
stealth61 [152]3 years ago
6 0

Answer:

1.33 cm          

Explanation:

Mass of copper, m=243 g

Side of the cube, a=3 cm

Volume of the copper cube, a^3=(3cm)^3=27cm^3

Density of copper, density =\frac{mass}{volume}

\rho=\frac{243gm}{27cm^3}=9gm/cm^3

Let the radius of the sphere be <em>r.</em>

Volume of the copper sphere, V=\frac{4}{3}\pi r^3

V=\frac{m}{\rho}\\\Rightarrow \frac{4}{3}\pi r^3= \frac{m}{\rho}\\\Rightarrow r =\sqrt[3]{ {\frac{3m}{4\pi \rho}}}

If the mass is 50 g, then the radius of the copper sphere is:

r=\sqrt[3]{\frac{3\times 50g}{4\pi 9g/cm^3}} =1.33 cm

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A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

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1) A plane's velocity increases from 40 m/s to 100 m/s over a 10 second interval. What is the plane's average acceleration for t
Yakvenalex [24]

Answer:

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Explanation:

Recall that the average acceleration (a)  is defined by the change in velocity from an initial velocity (v_i), to a final velocity (v_f) over the time (t) it took that change to happen. Then, in mathematical terms this is:

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with our information this becomes:

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3 years ago
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