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
Answer:
D
Explanation:
First we define our variables
V0=29.4
a=-9.8
V=0
We have to find the maximum displacement , which I will define as X
We use formula v^2=v0^2+2aX
All we do is substitute our values
0=29.4^2-19.6X
29.4^2=19.6X
X=29.4^2/19.6=44.1
Answer:
Invisible UV energy reacts with emulsion sensitizer and hardens the stencil so it won't dissolve with water and rinse down the drain
Explanation:
12.00 min = 0.2 hr
8.00 min = 0.15 hr
Total distance:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)
= 8.25 km
Average speed:
(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3
= 15 km/hr
Change in position:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)
= 0.25 km
Average velocity:
(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3
≈ 1.67 m/s
