Answer:
The work done is 205 kJ.
Explanation:
Hi there!
Work can be calculated using the following equation:
W = F · Δx
Where:
W = work
F = applied force
Δx = displacement
In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:
W = ∫ F · dx
F = 3.6 N/m³ · x³ - 76 N
W = ∫ (3.6 x³ - 76)dx
W = 0.9 x⁴ - 76x
Evaluating from xi to xf:
W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m
W = 205 kJ
Velocity, direction, or both at the same time.
D.) White Dwarf
It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.
Hope this helps!
Answer:
The average induced emf around the border of the circular region is 
.
Explanation:
Given that,
Radius of circular region, r = 1.5 mm
Initial magnetic field, B = 0
Final magnetic field, B' = 1.5 T
The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :
So, the average induced emf around the border of the circular region is .
1st <span>the total </span>energy<span> of an </span>isolated system<span> is constant; energy can be transformed from one form to another, but can be neither created nor destroyed. ▲U=Q-W
</span><span>
2nd the total </span>entropy<span> can never decrease over time for an </span>isolated system, that is, a system in which neither energy nor matter can enter nor leave.
DS (Greater than or equal to) 0
