Answer:
F = 19.375 x 10^-6 N
Explanation:
This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.
It is defined as
F = (ke * q1 * q2)/ r^2
Where,
ke = is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2
q1 = 5.0 x 10^-8 C
q2 = 1.0 x 10^-7 C
r = 5 ft = 1,524 m
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)
F = 19.375 x 10^-6 N
Answer:
a) a = 5.03x10¹³ m/s²
b)
Explanation:
a) The acceleration of the positron can be found as follows:
(1)
Also,
(2)
By entering equation (1) into (2), we have:
<u>Where:</u>
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
b) The positron's speed can be calculated using the following equation:
<u>Where</u>:
: is the final speed =?
: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
I hope it helps you!
Explanation:
Let x = distance of from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque about the fulcrum is zero:
Solving for <em>x</em>,
The correct answer would be A "<span>A light-year is the distance light travels in a year.
This is considered a unit of distance connected to the distance that light can travel in one year. It is proved that light travels at 300,000 km per second so, in 1 year, it might travel 10 trillion km.
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Answer:
Explanation:
Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.
In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.
Gauss' Law states that
Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.
So,