Practice india-cet-kcet questions with answers and explanations.
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KE=½mv²=32.0 J.
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V=IR=35 V.
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R=9 Ω.
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mgh=70 J.
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V=IR=15 V.
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By Vieta, α+β=6.
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Determinant=4×5=20.
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|v|²=5²+7²=74.
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f′(x)=2x, so f′(6)=12.
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(7+3)/2=5.0.
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s=vt=20 m.
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KE=½mv²=160.0 J.
