We are given:
Step 1: Use Newton’s Second Law to find acceleration:
\[ \vec{a} = \frac{\vec{F}}{m} = 2t \hat{i} + 3t^2 \hat{j} \]
Step 2: Integrate to find velocity:
\[ \vec{v} = \int \vec{a} \, dt = \int (2t \hat{i} + 3t^2 \hat{j}) \, dt = t^2 \hat{i} + t^3 \hat{j} \]
Step 3: Power is given by dot product \( \vec{F} \cdot \vec{v} \):
\[ P = (2t \hat{i} + 3t^2 \hat{j}) \cdot (t^2 \hat{i} + t^3 \hat{j}) = 2t^3 + 3t^5 \] \[ \Rightarrow P = 2t^3 + 3t^5 \, \text{W} \]
A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
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 ______