Practice india-cet-assam-cee questions with answers and explanations.
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R=12 Ω.
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mgh=20 J.
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s=vt=27 m.
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KE=½mv²=50.0 J.
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V=IR=40 V.
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KE=½mv²=9.0 J.
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By Vieta, α+β=5.
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Determinant=3×5=15.
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|v|²=4²+7²=65.
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f′(x)=2x, so f′(5)=10.
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(6+3)/2=4.5.
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s=vt=24 m.
