The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval,
:
First we need to find the net charge flowing at a certain point of the wire in one second,
. Using I=0.92 A and re-arranging the previous equation, we find
Now we know that each electron carries a charge of
, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
Answer:
Thin, aluminium and buried underground.
Explanation:
When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.
i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.
ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.
iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.
Answer:
A. 260 mm
B. - 18
C. 175
Explanation:
A
The expression for the lens equation is
v = 259 mm
= 260 mm or 26 cm (to 2 s.f)
check the attached files for additional solution
Explanation:
The x component of the resultant force is the sum of the x components of the individual forces.
Fₓ = 65.0 cos 30° − 20.0 sin 20° − 30.0
Fₓ = 19.5
The y component of the resultant force is the sum of the y components of the individual forces.
Fᵧ = 65.0 sin 30° − 20.0 cos 20°
Fᵧ = 13.7
The magnitude is found with Pythagorean theorem:
F² = Fₓ² + Fᵧ²
F² = (19.5)² + (13.7)²
F = 23.8
