Question 1
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Question 2
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Question 3
I. Area
II. Thrust
III. Pressure
IV. Mass
1. Area is derived from length multiplied by length (\( \text{m}^2 \)).
2. Thrust is a type of force, which is mass multiplied by acceleration (\( \text{kg} \cdot \text{m/s}^2 \) or Newtons).
3. Pressure is derived from force divided by area (\( \text{N/m}^2 \) or Pascals).
4. Mass is one of the basic fundamental quantities in the SI system and its unit (kg) is not derived from others. Since Area, Thrust, and Pressure are all built from fundamental units, only I, II, and III are derived quantities.
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Question 4
Let:
\( m_1 = 0.5 \, \text{kg} \)
\( u_1 = 10 \, \text{ms}^{-1} \)
\( m_2 = 0.5 \, \text{kg} \)
\( u_2 = 0 \, \text{ms}^{-1} \) (at rest)
\( v \) = common velocity \[ m_1u_1 + m_2u_2 = (m_1 + m_2)v \] Substituting the values: \[ (0.5 \times 10) + (0.5 \times 0) = (0.5 + 0.5)v \] \[ 5 + 0 = 1.0v \] \[ v = 5.0 \, \text{ms}^{-1} \] The momentum of the first ball is shared between the combined masses, resulting in a common velocity of 5.0ms\(^{-1}\).
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Question 5
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Question 6
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Question 7
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Question 8
\( Q_T = 88000 \, \text{J} \)
\( m = 0.5 \, \text{kg} \)
\( c_w = 4200 \, \text{Jkg}^{-1}\text{C}^{-1} \)
\( \Delta\theta = 10 - 0 = 10 \, ^\circ\text{C} \) \[ Q_1 = mc_w\Delta\theta \] \[ Q_1 = 0.5 \times 4200 \times 10 = 21000 \, \text{J} \] Now, find the heat used for the phase change (\( Q_2 \)): \[ Q_T = Q_1 + Q_2 \] \[ 88000 = 21000 + Q_2 \] \[ Q_2 = 88000 - 21000 = 67000 \, \text{J} \] Since \( Q_2 = mL_f \): \[ 67000 = 0.5 \times L_f \] \[ L_f = \frac{67000}{0.5} = 134000 \, \text{Jkg}^{-1} \] \[ L_f = 134.0 \, \text{kJkg}^{-1} \] The specific latent heat of fusion of ice is therefore 134 kJ/kg.
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Question 9
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Question 10
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