By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
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Answer:
ΔT= 11.94 °C
Explanation:
Given that
mass of water = 10 kh
Time t= 15 min
Heat lot from water = 400 KJ
Heat input to the water = 1 KW
Heat input the water= 1 x 15 x 60
=900 KJ
By heat balancing
Heat supply - heat rejected = Heat gain by water
As we know that heat capacity of water
Now by putting the values
900 - 400 = 10 x 4.187 x ΔT
So rise in temperature of water ΔT= 11.94 °C
Answer:
(a) 20 MHz
(b) 1.025 KW
(c) 3.33 ns
(d) 33 pF
Explanation:
(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>
(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>
(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>
(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>
Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
Capillary action occurs When the adhesion to the walls stronger than dirt cohesive forces between a liquid molecules. the head towards Capillery action will take water in a uniform circular is limited by surface tension and, of course, gravity.
