How many milliliters are in 3.2 L? (1000 mL = 1L)
2 answers:
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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
(a) Let's convert the final speed of the car in m/s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s: