Practice india-cet-tg-eapcet questions with answers and explanations.
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mgh=80 J.
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s=vt=63 m.
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KE=½mv²=9.0 J.
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V=IR=18 V.
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R=13 Ω.
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KE=½mv²=256.0 J.
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By Vieta, α+β=18.
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Determinant=10×3=30.
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|v|²=2²+5²=29.
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f′(x)=2x, so f′(3)=6.
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(4+9)/2=6.5.
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s=vt=8 m.
