24. दोन वर्तुळे परस्परांना बिंदू व E मध्ये छेदतात. बिंदू E मधून
काढलेली त्यांची सामाईक वृत्तछेदिका वर्तुळांना बिंदू B व D
मध्ये छेदते. बिंदू B व D मधून काढलेल्या स्पर्शिक एकमेकींना बिंदू C मध्ये छेदतात. सिद्ध करा : DABCD
चक्रीय आहे.
Answers
Step-by-step explanation:
\begin{gathered}\Large\bf{\color{peru}GiVeN,} \\ \end{gathered}
GiVeN,
A force of \bf{(5\hat{i}\:+\:4\hat{j})\:N}(5
i
^
+4
j
^
)N displaces a particle through \bf{(3\hat{i}\:-\:3\hat{j})\:m}(3
i
^
−3
j
^
)m .
\begin{gathered}\longmapsto\:\:\bf\blue{Force\:(F)\:=\:(5\hat{i}\:+\:4\hat{j})\:N} \\ \end{gathered}
⟼Force(F)=(5
i
^
+4
j
^
)N
\begin{gathered}\longmapsto\:\:\bf{\color{lime}Displacement\:(s)\:=\:(3\hat{i}\:-\:3\hat{j})\:m} \\ \end{gathered}
⟼Displacement(s)=(3
i
^
−3
j
^
)m
Power (P) = 0.6 W
\begin{gathered}\bf{\color{indigo}We\:know\:that,} \\ \end{gathered}
Weknowthat,
\begin{gathered}\red\bigstar\:\:{\pink{\boxed{\boxed{\bf{\color{cyan}Work\:done\:=\:F\:.\:s\:}}}}}---(i) \\ \end{gathered}
★
Workdone=F.s
−−−(i)
\begin{gathered}\Large\bf\orange{Or} \\ \end{gathered}
Or
\begin{gathered}\purple\bigstar\:\:{\green{\boxed{\boxed{\bf{\color{olive}Work\:done\:=\:Power\times{time}\:}}}}}--(ii) \\ \end{gathered}
★
Workdone=Power×time
−−(ii)
\begin{gathered}\bf\red{From\:equ^n(i)\:\&\:(ii)\:we\:get,} \\ \end{gathered}
Fromequ
n
(i)&(ii)weget,
\begin{gathered}:\implies\:\:\bf{F\:.\:s\:=\:Power\times{time}\:} \\ \end{gathered}
:⟹F.s=Power×time
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{F\:.\:s}{Power}\:} \\ \end{gathered}
:⟹Time=
Power
F.s
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{(5\hat{i}\:+\:4\hat{j})\:.\:(3\hat{i}\:-\:3\hat{j})}{0.6}\:} \\ \end{gathered}
:⟹Time=
0.6
(5
i
^
+4
j
^
).(3
i
^
−3
j
^
)
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{15\:(\hat{i}.\hat{i}\:+\:12\:(\hat{j}.\hat{i})\:-\:15\:(\hat{i}.\hat{j})\:-\:12\:(\hat{j}.\hat{j})}{0.6}\:} \\ \end{gathered}
:⟹Time=
0.6
15(
i
^
.
i
^
+12(
j
^
.
i
^
)−15(
i
^
.
j
^
)−12(
j
^
.
j
^
)
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{15\times{1}\:+\:12\times{0}\:-\:15\times{0}\:-\:12\times{1}}{0.6}\:} \\ \end{gathered}
:⟹Time=
0.6
15×1+12×0−15×0−12×1
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{15\:-\:12}{0.6}\:} \\ \end{gathered}
:⟹Time=
0.6
15−12
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{3}{0.6}\:} \\ \end{gathered}
:⟹Time=
0.6
3
\begin{gathered}:\implies\:\:\bf{Time\:=\:\dfrac{30}{6}\:} \\ \end{gathered}
:⟹Time=
6
30
\begin{gathered}:\implies\:\:\bf\purple{Time\:=\:5\:s\:} \\ \end{gathered}
:⟹Time=5s
\Large\bold\therefore∴ The time of action is 5 seconds.