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kifflom [539]
3 years ago
11

A ball has an electric charge of +1.5 × 10-9 coulombs. At what distance from the ball's center is the electric field strength eq

ual to 5.8 × 105 newtons/coulomb?
Physics
1 answer:
Veronika [31]3 years ago
3 0
F = kq1q2/r<span>2
Where,
F - Coulomb Force
k - constant value which is equal to </span>8.98 × 10^9<span> newton square metre per square coulomb
q1 and q2 -  two electric charges
r - distance.

5.8 * 10^5 = 1.5 * 10^-9 / r^2
</span><span>5.8 * 10^5 r^2 = 1.5*10^-9
</span>r^2 = 0.0000258620
r = 0.0050854694

So the distance is equal to 5.09 x 10^-3 
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2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m

 
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Explanation:

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3 years ago
What are tectonic plates, and which of Earht's layers are they composed of?
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3 years ago
Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.
mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, \rho=10^{-6}\ \Omega -m

Density, d=8.31\times 10^3\ kg/m^3

(A) The length of the wire is given by using the definition of resistance as :

Volume,

V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3

Area,

A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}....(1)

R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m

(b)  Equation (1) becomes :

A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m

Hence, this is the required solution.                                                                  

5 0
3 years ago
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