Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity 
= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4
Use this formula to find your answer...
Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms
frequency (f)=1/( Time period).
Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.
Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.
Folk song
(Word cap filler)
Answer: The HUMAN EYE
Explanation:
The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.
In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.
The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.
The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.
Answer:
Solving for time :
(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known
2. The unknown constant that we want to solve)
s = (1/2)(u+v)t <--- one of the formulas
from linear motion
s (distance) = 0.05m
u (initial velocity) = 100m/s
v (final velocity) = 0 m/s (it stops)
t (time taken for change in velocity) = to be found
0.05 = (1/2)(100+0)t
t = 0.001 seconds
Solving for the resistant force :
Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.
When the bullet stops :
F net = 0
F r = F imp
F r = (mu -mv)/t
F r = (0.01x100-0.01x0)/0.001
F r = 1/0.001
F r = 1000N
