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3 years ago

8

How many electrons must be removed from each of two 5.69-kg copper spheres to make the electric force of repulsion between them

equal in magnitude to the gravitational attraction between them?

1 answer:

CaHeK987 [17]3 years ago

7 0

Answer:

$n=3.056*10^{9} Electrons$

Explanation:

Please see attached file

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You are In-line skates at the top of a small hill. Your potential energy is equal to 1000 J. The last time we checked, your mass

stira [4]

A) <u>Weight = mass × acceleration (due to gravity) </u>

= 60×9.8

= 588 N

<u>B) Potential energy = mass x gravity x change in height </u>

1,000 = 60.0 x 9.8 x h

h = 1.7 m

<u>C) Kinetic energyF = potential energyI </u>

KEF = 1/2mv2

PEI = mgh = 1,000 J

1/2mv2 = 1,000

1/2(60.0)v2 = 1,000

v2 = 33.33

v = 5.77 m/s

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4 years ago

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

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*Materials that regulate the flow of current through them *

4vir4ik [10]

Answer:

electromagnet

Explanation:

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3 years ago

To pop a balloon you stab it with a pencil. If the area of the pencil tip is .01 cm² and the pressure applied by the pencil to t

Romashka-Z-Leto [24]

Answer:

Push with force of 1N

Explanation:

I have explained in the paper.

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3 0

3 years ago

A spherical helium filled balloon (B) with a hanging passenger cage being held by a single vertical cable (C) attached to Earth

Gre4nikov [31]

Answer:

The tension is $T = 4326.7 \ N$

Explanation:

From the question we are told that

The total mass is $m = 200 \ kg$

The radius is $r = 5 \ m$

The density of air is $\rho_a = 1.225 \ kg/m^3$

Generally the upward force acting on the balloon is mathematically represented as

$F_N = T + mg$

=> $(\rho_a * V * g ) = T + mg$

=> $T = (\rho_a * V * g ) - mg$

Here V is the volume of the spherical helium filled balloon which is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

=> $V = \frac{4}{3} * 3.142 *(5)^3$

=> $V = 523.67\ m^3$

So

$T = (1.225 * 523.67* 9.8 ) - 200 * 9.8$

$T = 4326.7 \ N$

5 0

3 years ago