If the resulting trajectory of the charged particle is a circle,what is , the angular frequency of the circular motion?Express i
1 answer:
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Answer:
Option A = 1.
Explanation:
So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;
(1). First, Both objects A and D represent fixed.
(2). Both objects A and D are negatively-charged particles of equal magnitude.
(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."
(4). "Object C shows a moving, positively-charged particle."
So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.
We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.
The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).
The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i =
where R is the resistance
R =
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i =
q =
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
Answer:
Explanation:
The gravitational potential at a point on the Earth surface is given by:
where
G=6.67×10^-11Nm^2kg^-2 is the gravitational constant
M=5.98×10^24kg is the Earth's mass
R=6.38×10^6 m is the Earth's radius
Substituting the numbers into the equation, we find