what is the minimum heat if A 500cm³ of water is to be heated from a room temperature (28°C) to 100°C in order to prepare hot cu
1 answer:
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Answer:
R_total = 14.57 Ω , V_C = 1.176 V
Explanation:
To solve this circuit we are going to find the equivalent resistance of each branch, let's remember
* Serial resistance
= ∑
* For resistance in parallel
1 / R_{eq} = ∑ 1/R_{i}
We solve the two branches of the wheatstone bridge
Series resistors
Branch B
R_B = Rb + R4
R_B = 2 + 18
R_B = 20 Ω
Branch C
R_C5 = Rc + R5
R_C5 = 3 + 12
R_C5 = 15 Ω
Resistance in parallel R_B and R_C5
1 / R_BC = 1 / R_B + 1 / R_C5
1 / R_BC = 1/20 + 1/15 = 0.116666
R_BC = 8.57 Ω
Now we have a single branch, we solve the series resistance
R_total = R_A + R_BC
R_total = 6 + 8.57
R_total = 14.57 Ω
b) they ask us for the voltage in the resistance R_C
Let's remember that the voltage in a series circuit is the sum of the voltages
10 = V_a + V_BC
10 = i R_a + i R_BC = i (R_a + R_BC)
i = 10 / (R_a + R_BC)
i = 10 / (14.57)
i = 0.6863 A
The current in the series circuit is constant
V_BC = i R_BC
V_BC = 0.6863 8.57
V_BC = 5.8819 V
This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.
Branch C
V_BC = i R_C5
i = V_BC / R_C5
i = 5.8819 / 15
i = 0.39213 A
In this branch we have two resistors in series, let's remember that the current in a series circuit is constant
V_C = i R_C
V_C = 0.39213 3
V_C = 1.176 V
Answer:
(a) Elongation of rod = 0.19732 mm
(b) Change in diameter = 0.00651 mm
Explanation:
Circular area at end of steel rod = pi * diameter^2 / 4
Area = 3.801 * 10^(-4) meter squared
Stress = force / area
Stress = 75000 / (3.801 * 10^-4)
Stress = 197316495 Pa OR 0.197 GPa
Modulus of elasticity = stress / strain
200 = 0.19732 / Strain
Strain = 0.0009866 (longitudinal)
(a) Strain = change in length / initial length
0.0009866 = Elongation of rod / 200
Elongation of rod = 0.19732 mm
(b) Poisson ratio = lateral strain / longitudinal strain
0.3 = lateral strain / 0.0009866
Lateral strain = 0.000296
Lateral strain = Change in diameter / original diameter
0.000296 = Change in diameter / 22
Change in diameter = 0.00651 mm
Answer:
T = 692.42 N
Explanation:
Given that,
Mass of hammer, m = 8.71 kg
Length of the chain to which an athlete whirls the hammer, r = 1.5 m
The angular sped of the hammer,
We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :
So, the tension in the chain is 692.42 N.