Example Question - equilibrium
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Cost Schedule Analysis for a Firm
(a) To calculate the Total Variable Cost (TVC) and Total Cost (TC): <p></p> <p>TVC = Output \times Marginal Cost</p> <p>TC = Total Fixed Cost + TVC</p> <p>At output 1: TVC = 1 \times 10 = 10, TC = 20 + 10 = 30</p> <p>At output 2: TVC = 2 \times 20 = 40, TC = 20 + 40 = 60</p> <p>At output 3: TVC = 3 \times 30 = 90, TC = 20 + 90 = 110</p> <p>At output 4: TVC = 4 \times 40 = 160, TC = 20 + 160 = 180</p> <p></p> (b) To calculate the Total Revenue (TR) at each level of output if price is $30: <p></p> <p>TR = Price \times Output</p> <p>At output 1: TR = 30 \times 1 = 30</p> <p>At output 2: TR = 30 \times 2 = 60</p> <p>At output 3: TR = 30 \times 3 = 90</p> <p>At output 4: TR = 30 \times 4 = 120</p> <p></p> (c) To calculate the profit/loss at each level of output: <p></p> <p>Profit/Loss = TR - TC</p> <p>At output 1: Profit/Loss = 30 - 30 = 0</p> <p>At output 2: Profit/Loss = 60 - 60 = 0</p> <p>At output 3: Profit/Loss = 90 - 110 = -20 (loss)</p> <p>At output 4: Profit/Loss = 120 - 180 = -60 (loss)</p> <p></p> (d) The firm will be in equilibrium at the output level where TR = TC: <p></p> <p>At output 1 and 2 the TR equals TC, so the equilibrium price would be at output levels 1 and 2 if the price should remain at $30.</p> <p></p> (e) With the provided information, it's not possible to determine the time period of the firm's operation, as it requires additional business cycle data or context.
Finding Equilibrium Force and Fulcrum Reaction in a Leverage System
Given: Mass, \( m = 72 \text{ kg} \) Distance to fulcrum, \( b = 12 \text{ cm} = 0.12 \text{ m} \) Fulcrum reaction, \( R = 807.2 \text{ N} \) Gravity, \( g = 9.81 \text{ m/s}^2 \) Total length of lever, \( L = 96 \text{ cm} = 0.96 \text{ m} \) Distance \( a = L - b = 0.96 \text{ m} - 0.12 \text{ m} = 0.84 \text{ m} \) To find the equilibrium force \( F \), we use the principle of moments (torque \( \tau = Fd \)), where \( d \) is the distance from the fulcrum: <p>\( \tau_{\text{clockwise}} = \tau_{\text{counterclockwise}} \)</p> <p>\( F \cdot a = m \cdot g \cdot b \)</p> <p>\( F \cdot 0.84 \text{ m} = 72 \text{ kg} \cdot 9.81 \text{ m/s}^2 \cdot 0.12 \text{ m} \)</p> <p>\( F = \frac{72 \cdot 9.81 \cdot 0.12}{0.84} \)</p> <p>\( F = \frac{847.872}{0.84} \)</p> <p>\( F \approx 1009.61 \text{ N} \)</p> To confirm that the given \( F = 100.9 \text{ N} \) is not correct, and the correct \( F \) is approximately \( 1009.61 \text{ N} \).
Torque Balance in Lever System
<p>Given that the mass \( m = 72 \) kg, the distance from the fulcrum to the mass \( b = 12 \) cm, and the total length of the lever \( L = 96 \) cm, we need to find the force \( F \) to balance the lever and the reaction \( R \) at the fulcrum.</p> <p>To find the force \( F \), we use the principle of moments (torque balance) where the clockwise moments equal the anticlockwise moments about the pivot.</p> <p>The weight of the mass \( W \) acting at a distance \( b \) from the fulcrum is given by:</p> <p>\( W = m \cdot g \)</p> <p>\( W = 72 \cdot 9.8 \) (taking \( g = 9.8 \, \text{m/s}^2 \))</p> <p>\( W = 705.6 \) N</p> <p>The anticlockwise moment due to \( W \) is \( W \cdot b \).</p> <p>Let \( a = L - b \). Then \( a = 96 - 12 = 84 \) cm \( = 0.84 \) m.</p> <p>The clockwise moment due to \( F \) is \( F \cdot a \).</p> <p>Setting the moments equal for balance:</p> <p>\( W \cdot b = F \cdot a \)</p> <p>\( F = \frac{W \cdot b}{a} \)</p> <p>\( F = \frac{705.6 \cdot 0.12}{0.84} \)</p> <p>\( F = 100.8 \) N</p> <p>Now, to find the reaction \( R \) at the fulcrum, we use equilibrium of vertical forces:</p> <p>\( R = W + F \)</p> <p>\( R = 705.6 + 100.8 \)</p> <p>\( R = 806.4 \) N</p> <p>Since \( F \) is 100.8 N which is approximately 100.9 N, and \( R \) is 806.4 N which is approximately 806.7 N.</p>
Understanding the Direction of Water Movement in Osmosis
<p>La ósmosis es el movimiento del agua desde un lugar de _______ concentración de solutos hasta un lugar de _______ concentración de solutos; por tanto, el agua en este ejemplo se movería hacia dentro de la célula hasta que se alcance el equilibrio.</p> <p>La primera palabra que falta es "baja" y la segunda palabra que falta es "alta". Así que, la afirmación completa sería:</p> <p>La ósmosis es el movimiento del agua desde un lugar de baja concentración de solutos hasta un lugar de alta concentración de solutos; por tanto, el agua en este ejemplo se movería hacia dentro de la célula hasta que se alcance el equilibrio.</p>
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