find the third proportional to 1m 60cm,40cm
Answers
Step-by-step explanation:
GIVEN:−
A body has displacement (2, 4, -6) to (6, -4, 4) under a constant force \sf{2\hat{i}+3\hat{j}-\hat{k}}2i^+3j^−k^ .
\bigstar\ \sf{\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}}★ F=2i^+3j^−k^
\large \underline{\underline{\sf{\color{orange}{TO\ FIND:-}}}}TO FIND:−
The work done by the body.
\large \underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}SOLUTION:−
According to the question,
\bullet\ \sf{\overrightarrow{r_1}=2\hat{i}+4\hat{j}-6\hat{k}}∙ r1=2i^+4j^−6k^
\bullet\ \sf{\overrightarrow{r_2}=6\hat{i}-4\hat{j}+4\hat{k}}∙ r2=6i^−4j^+4k^
\sf{\to \Delta \overrightarrow{r}=\overrightarrow{D}=\overrightarrow {r_2}-\overrightarrow {r_1}}→Δr=D=r2−r1
Solving further :-
\longrightarrow \sf{\Delta \overrightarrow{r}= 6\hat{i}-4\hat{j}+4\hat{k}-(2\hat{i}+4\hat{j}-6\hat{k}) }⟶Δr=6i^−4j^+4k^−(2i^+4j^−6k^)
\longrightarrow \sf{\Delta \overrightarrow{r}= 6\hat{i}-4\hat{j}+4\hat{k}-2\hat{i}-4\hat{j}+6\hat{k}}⟶Δr=6i^−4j^+4k^−2i^−4j^+6k^
\longrightarrow \sf{\Delta \overrightarrow{r}= 4\hat{i}-8\hat{j}+10\hat{k}}⟶Δr=4i^−8j^+10k^
__________________
We are given with :
\dag\ \sf{\overrightarrow{F}=2\hat{i}+3\hat{j}-\hat{k}}† F=2i^+3j^−k^
We know that,
\sf{Work\ done,\ W=\overrightarrow{F}.\overrightarrow{D}}Work done, W=F.D
Solving further :-
\sf{\longrightarrow W= (2\hat{i}+3\hat{j}-\hat{k}).(4\hat{i}-8\hat{j}+10\hat{k})}⟶W=(2i^+3j^−k^).(4i^−8j^+10k^)
\sf{\longrightarrow W= 8-24-10}⟶W=8−24−10
\boxed{\bf{\longrightarrow W= -26\ J}}⟶W=−26 J
OR
\boxed{\bf{\longrightarrow W= -26\ N.m}}⟶W=−26 N.m
Therefore, the work done by the body is -26 Joules or -26 Newton·meter.
Step-by-step explanation:
1st proportion is 1m =100cm
2nd proportion is 60cm
3rd proportion is 40cm
100:60=X:40
100/60=X/40
X=10/60*40
X=5/3"40X=1.66*40
X=66.4
hope it helps........