Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J
From the first Law of thermodynamics, obtain useful work performed as
W = Qin - Qout
= 0.5 x 10⁵ J
By definition, the efficiency is
η = W/Qin
= 100*(0.5 x 10⁵/4 x 10⁵)
= 12.5%
Answer: The efficiency is 12.5%
Answer:
(1) tropical storm
(2) Severe tropical storm
(3) 305 kmh
Answer:
1.1775 x 10^-3 m^3 /s
Explanation:
viscosity, η = 0.250 Ns/m^2
radius, r = 5 mm = 5 x 10^-3 m
length, l = 25 cm = 0.25 m
Pressure, P = 300 kPa = 300000 Pa
According to the Poisuellie's formula
Volume flow per unit time is
V = 1.1775 x 10^-3 m^3 /s
Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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