Because each case i observed happened due to the experiments that I did with my partner.
R= (rou * L) / area
where R is the wire resistance
rou: resistivity of the wire material
L : wire length
A : cross section area of wire
by sub.
0.757= (rou*25)/ 3.5*10^-6
25*rou = 2.6495*10^-6
rou= 1.0598*10^-7 ohm.m
Answer:
592000 J
Explanation:
We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:
1 Pa = 1 Kg/ms²
Therefore,
3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²
Next, we shall determine the workdone.
Workdone is given by the following equation:
Workdone (Wd) = pressure (P) × change in volume (ΔV)
Wd = PΔV
With the above formula, the work done can be obtained as follow:
Pressure (P) = 3.7×10⁵ Kg/ms²
Change in volume (ΔV) = 1.6 m³
Workdone (Wd) =?
Wd = PΔV
Wd = 3.7×10⁵ × 1.6
Wd = 592000 Kgm²/s²
Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:
1 Kgm²/s² = 1 J
Therefore,
592000 Kgm²/s² = 592000 J
Therefore, the Workdone is 592000 J.
Answer:
11.94
Explanation:
Remark
Find the Potential Energy at the top.
Givens
m = 65 kg
h = 16.2 m
g = 9.81
PE = 65 * 9.81 * 16.2
PE = 10329.93
The tricky part is what do you do about Friction?
Formula
PE = Friction + KE
Solution
PE = 10329.93 Joules
Friction = 5700 Joules
Find the KE
10329.93 = 5700 + KE
KE = 10329.93 - 5700
KE = 4629.93
Find V from the KE formula
KE = 4629.93
m = 65
KE = 1/2 m v^2
KE = 1/2 65 v^2
4629.93 = 1/2 65 v^2
v^2 = 142.46
v = √142.46
v = 11.94
F=m•a(squared)
F=8,000•4(squared)
F=8,000•16
F=128,000
then divide by 2 which is
64,000 Joules
