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A force $\vec{F} = 6t^2\,\hat{i} + 4t\,\hat{j}$ acts on a particle of mass $3\text{ kg}$, initially at rest. Find its velocity at $t = 3\text{ s}$.
Acceleration: $\vec{a} = \dfrac{\vec{F}}{m} = 2t^2\,\hat{i} + \dfrac{4t}{3}\,\hat{j}$
Velocity: $\vec{v} = \displaystyle\int_0^3 \vec{a}\,dt = \left[\dfrac{2t^3}{3}\,\hat{i} + \dfrac{2t^2}{3}\,\hat{j}\right]_0^3 = \dfrac{2(27)}{3}\,\hat{i} + \dfrac{2(9)}{3}\,\hat{j} = \mathbf{18\hat{i} + 6\hat{j}}$
Vehicle X cost $80,000, useful life 5 years, residual value $5,000, depreciated straight-line. After 3 years, part-exchange value was $20,000. What was the loss on disposal?
Option B ($15,000) is correct.
Annual depreciation = ($80,000 − $5,000) ÷ 5 = $15,000
Accumulated depreciation after 3 years = $45,000
Carrying value = $80,000 − $45,000 = $35,000
Loss on disposal = $35,000 − $20,000 = $15,000
The ratio $\dfrac{1}{3} : \dfrac{3}{8}$ is equal to which of the following ratios?
To find the ratio $\dfrac{1}{3} : \dfrac{3}{8}$, divide the first by the second:
$\dfrac{1/3}{3/8} = \dfrac{1}{3} \times \dfrac{8}{3} = \dfrac{8}{9}$
So the ratio is $\dfrac{8}{9}$, which means $8$ to $9$.
You can verify: multiply both original fractions by 24 (the LCM of 3 and 8): $\dfrac{1}{3} \times 24 = 8$ and $\dfrac{3}{8} \times 24 = 9$. Ratio is $8:9$.
From the definition of Bulk Modulus $K$:
$\frac{\Delta V}{V} = \frac{P}{K}$ (Volume strain due to pressure)
From thermal expansion, the increase in volume is:
$\frac{\Delta V}{V} = \gamma \Delta T = 3\alpha \Delta T$
To restore the original size, the thermal expansion must equal the pressure compression:
$3\alpha \Delta T = \frac{P}{K} \implies \Delta T = \frac{P}{3\alpha K}$
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