I changed my undershorts. The elastic on the old ones I put on that day was deteriorated, and it completely failed when I dripped lab coffee on it, causing falldown.
Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire, 
= 
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density, 
where,
=Avogadro's Number
Now,
the drift velocity,
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
Hello!
Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:
F = 5 · 20
F = 100N
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
