A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.
<h3>What is electric potential energy?</h3>
The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.
The electric potential energy is given by the relation,
V = kQ/r
where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.
Substitute the values into the expression to get the distance between the charges.
8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r
r =3.34 x 10⁷ m
Thus, the distance between the charges will be 3.34 x 10⁷ m.
Learn more about electric potential energy.
brainly.com/question/12645463
#SPJ1
Answer:
Potential energy only
Explanation:
at the top of its swing the pendulum stops moving , (therefore it has no KINETIC energy) thus all of the energy is stored as potential energy.
Answer:
Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...
Explanation:
Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.
For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc
If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc
Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...
Explanation:
Hey there!
Given;
Force (f) = 100 N
Acceleration (a) = 10m/s^2.
Mass (m)=?
We have;
Force = mass * acceleration.
So,
100 = m * 10
Or, 100 = 10m
Or, m= 100/10
Therefore, mass is 10kg.
<em><u>Hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω
Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂
The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁
Hence, the protostar will be rotating 16 </span><span>times faster.</span>