A force of 100N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 12 meters. How much is the work done?
Answers
Answer:
\large\underline{\sf{Solution-}}Solution−
Given that,
Radius of driving wheel of engine = 1.75 m
Distance covered = 22 km = 22000 m
Let assume that number of revolutions be n.
We know,
Distance covered = Number of revolutions × Circumference of wheel.
So, it means
\begin{gathered}\rm \: n \times 2 \times \pi \times r \: = \: 22000 \\ \end{gathered}n×2×π×r=22000
\begin{gathered}\rm \: n \times \dfrac{22}{7} \times 1.75 = \: 11000 \\ \end{gathered}n×722×1.75=11000
\begin{gathered}\rm \: n \times \dfrac{22}{7} \times \dfrac{175}{100} = \: 11000 \\ \end{gathered}n×722×100175=11000
\begin{gathered}\rm \: n \times 22 \times 25 = 1100000 \\ \end{gathered}n×22×25=1100000
\begin{gathered}\rm \: n \times 25 = 50000 \\ \end{gathered}n×25=50000
\begin{gathered}\bf\implies \: \: n \: = \: 2000 \\ \\ \end{gathered}⟹n=2000
Thus, it make 2000 revolutions to cover distance of 22 km.
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{ \red{ \mathfrak{Additional\:Information}}}AdditionalInformation
\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}†FormulasofAreas:−⋆Square=(side)2⋆Rectangle=Length×Breadth⋆Triangle=21×Base×Height⋆Scalene△=s(s−a)(s−b)(s−c)⋆Rhombus=21×d1×d2⋆Rhombus=21d4a2−d2⋆Parallelogram=Base×Height⋆Trapezium=21(a+b)×Height⋆EquilateralTriangle=43(side)2