Work

The total resistance of the wire is $9%20%5C%2C%20%5COmega$ , and it is bent to form an equilateral triangle. The resistance of each side of the triangle is: $R_%7B%7Bside%7D%7D%20%3D%20%5Cfrac%7B9%7D%7B3%7D%20%3D%203%20%5C%2C%20%5COmega$ When two vertices are connected, the equivalent resistance across the two vertices is found by considering the resistances in parallel.

One side between the vertices is $3%20%5C%2C%20%5COmega$ , and the other two sides are in series: $R_%7B%7Bseries%7D%7D%20%3D%203%20%2B%203%20%3D%206%20%5C%2C%20%5COmega$ Now, the two resistances $3%20%5C%2C%20%5COmega$ and $6%20%5C%2C%20%5COmega$ are in parallel. The equivalent resistance $R_%7B%7Beq%7D%7D$ is: $%5Cfrac%7B1%7D%7BR_%7B%7Beq%7D%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B6%7D%20%3D%20%5Cfrac%7B3%7D%7B6%7D$ $R_%7B%7Beq%7D%7D%20%3D%20%5Cfrac%7B6%7D%7B3%7D%20%3D%202%20%5C%2C%20%5COmega$ Thus, the equivalent resistance across any two vertices is $%5Cboxed%7B2%20%5C%2C%20%5COmega%7D$ .

The problem involves calculating the work done by a resistive force (like air resistance) on a falling object. We can use the Work-Energy Theorem, which states that the total work done by all forces acting on an object is equal to the change in its kinetic energy.
$

W_%7B%5Ctext%7Btotal%7D%7D%20%3D%20%5CDelta%20K

%3Cbr%3EThe%20forces%20acting%20on%20the%20raindrop%20are%20gravity%20and%20the%20resistive%20force.%20Therefore%3A%3Cbr%3E

W_%7B%5Ctext%7Bgravity%7D%7D%20%2B%20W_%7B%5Ctext%7Bresistive%7D%7D%20%3D%20K_%7B%5Ctext%7Bfinal%7D%7D%20-%20K_%7B%5Ctext%7Binitial%7D%7D

%3Cbr%3EStep%201%3A%20Identify%20given%20values%20in%20SI%20units.%3Cbr%3EMass

m = 1

g

= 10^{-3}

kg%3Cbr%3EInitial%20velocity

u = 0

m%2Fs%3Cbr%3EFinal%20velocity

v = 5

m%2Fs%3Cbr%3EHeight

h = 1

km

= 1000

m%3Cbr%3EAcceleration%20due%20to%20gravity

g = 10

m%2Fs

%3Cbr%3EStep%202%3A%20Calculate%20work%20done%20by%20gravity.%3Cbr%3E

W_%7B%5Ctext%7Bgravity%7D%7D%20%3D%20mgh%20%3D%20(10%5E%7B-3%7D)%20%5Ctimes%20(10)%20%5Ctimes%20(1000)%20%3D%2010%20%5Ctext%7B%20J%7D

%3Cbr%3EStep%203%3A%20Calculate%20the%20change%20in%20kinetic%20energy.%3Cbr%3E

K_%7B%5Ctext%7Binitial%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20u%5E2%20%3D%200

%3Cbr%3E

K_%7B%5Ctext%7Bfinal%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20(5)%5E2%20%3D%200.5%20%5Ctimes%200.001%20%5Ctimes%2025%20%3D%200.0125%20%5Ctext%7B%20J%7D

%3Cbr%3E

%5CDelta%20K%20%3D%200.0125%20-%200%20%3D%200.0125%20%5Ctext%7B%20J%7D

%3Cbr%3EStep%204%3A%20Solve%20for

W_{\text{resistive}}

.%3Cbr%3E

10%20%2B%20W_%7B%5Ctext%7Bresistive%7D%7D%20%3D%200.0125

%3Cbr%3E

W_%7B%5Ctext%7Bresistive%7D%7D%20%3D%200.0125%20-%2010%20%3D%20-9.9875%20%5Ctext%7B%20J%7D

%3Cbr%3ERounding%20to%20two%20decimal%20places%2C%20we%20get

-9.98$ J.

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Work

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs