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what is linear co-efficient of linear expansion of rod if it's found to be 100m long at 20 degrees Celsius and 100.14m long at 100 degree Celsius??? |
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Answer» Answer: We will divide the Projectile MOTION into 2 simultaneously occuring 1D motions. Along X axis : \SF{ \therefore \: x = u \cos( \theta) \times t}∴x=ucos(θ)×t \sf{ \implies \: x = 20 \cos( 60 \DEGREE) \times 1}⟹x=20cos(60°)×1 \sf{ \implies \: x = 20 \times \dfrac{1}{2} \times 1}⟹x=20× 2 1
×1 \sf{ \implies \: x = 10 \: m}⟹x=10m Along Y axis : \sf{ \therefore \: y = u \sin( \theta) t - \frac{1}{2} g {t}^{2}}∴y=usin(θ)t− 2 1
gt 2
\sf{ \implies \: y = 20 \sin( 60 \degree) \times 1- ( \frac{1}{2} \times 10 \times {1}^{2}} )⟹y=20sin(60°)×1−( 2 1
×10×1 2 ) \sf{ \implies \: y = 20 \times \frac{ \sqrt{3} }{2} - 5}⟹y=20× 2 3
−5 \sf{ \implies \: y =17.32 - 5}⟹y=17.32−5 \sf{ \implies \: y =12.32 \: m}⟹y=12.32m So , COORDINATE will be : \boxed{ \red{ \BOLD{ \sf{ \huge{(x,y) = (10,12.32)}}}}} (x,y)=(10,12.32)
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