Answer:
(I). The resistance of the copper wire is 0.0742 Ω.
(II). The resistance of the carbon piece is 1.75 Ω.
Explanation:
Given that,
Length of copper wire = 1.70 m
Diameter = 0.700 mm
Length of carbon piece = 20.0 cm
Cross section area
(I). We need to calculate the area of copper wire
Using formula of area


We need to calculate the resistance
Using formula of resistance

Put the value into the formula


(II). We need to calculate the resistance
Using formula of resistance

Put the value into the formula


Hence, (I). The resistance of the copper wire is 0.0742 Ω.
(II). The resistance of the carbon piece is 1.75 Ω.
Answer:
Technician A is right. The situation will happens even with only two bulbs in series
Explanation:
We must take into account that
1.- All electric device need its nominal voltage to operate
2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)
3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).
4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.
For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage
Answer:
See the answers below
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
<u>First case</u>
Vf = 6 [m/s]
Vo = 2 [m/s]
t = 2 [s]
![6=2+a*2\\4=2*a\\a=2[m/s^{2} ]](https://tex.z-dn.net/?f=6%3D2%2Ba%2A2%5C%5C4%3D2%2Aa%5C%5Ca%3D2%5Bm%2Fs%5E%7B2%7D%20%5D)
<u>Second case</u>
Vf = 25 [m/s]
Vo = 5 [m/s]
a = 2 [m/s²]
![25=5+2*t\\t = 10 [s]](https://tex.z-dn.net/?f=25%3D5%2B2%2At%5C%5Ct%20%3D%2010%20%5Bs%5D)
<u>Third case</u>
Vo =4 [m/s]
a = 10 [m/s²]
t = 2 [s]
![v_{f}=4+10*2\\v_{f}=24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D4%2B10%2A2%5C%5Cv_%7Bf%7D%3D24%20%5Bm%2Fs%5D)
<u>Fourth Case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
![v_{f}=5+8*10\\v_{f}=85 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D5%2B8%2A10%5C%5Cv_%7Bf%7D%3D85%20%5Bm%2Fs%5D)
<u>Fifth case</u>
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
