The correct answer is (C) : mk2r2t
\(a_r = k²rt² = \frac{v²}{r}\)
\(⇒ v² = k²r²t² \) or \( v = krt\)
and
\(\frac{d |v|}{dt} = kr\)
\(⇒ a_t = kr\)
\(⇒ | \stackrel{→}{F} . \stackrel{→}{v} | = (mkr) ( krt)\)
\(= mk^2r^2t \)= power delivered
0.5 g of an organic compound on combustion gave 1.46 g of $ CO_2 $ and 0.9 g of $ H_2O $. The percentage of carbon in the compound is ______ (Nearest integer) $\text{(Given : Molar mass (in g mol}^{-1}\text{ C : 12, H : 1, O : 16})$
Given: $ \Delta H_f^0 [C(graphite)] = 710 $ kJ mol⁻¹ $ \Delta_c H^0 = 414 $ kJ mol⁻¹ $ \Delta_{H-H}^0 = 436 $ kJ mol⁻¹ $ \Delta_{C-H}^0 = 611 $ kJ mol⁻¹
The \(\Delta H_{C=C}^0 \text{ for }CH_2=CH_2 \text{ is }\) _____\(\text{ kJ mol}^{-1} \text{ (nearest integer value)}\)
Consider the following reactions $ A + HCl + H_2SO_4 \rightarrow CrO_2Cl_2$ + Side Products Little amount $ CrO_2Cl_2(vapour) + NaOH \rightarrow B + NaCl + H_2O $ $ B + H^+ \rightarrow C + H_2O $ The number of terminal 'O' present in the compound 'C' is ______
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