Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
Answer:
the ball gains speed because of the force of the earth pulling on it and the kinetic energy coming ffrom the potental energy
Explanation:
Answer:
67.89%
Explanation:
Energy Efficiency=
Where
is useful energy and
is the total energy
The useful energy= (19-5-1.1) million= 12.9 million
The total energy= 19 million

Therefore, the efficiency of the boiler is 67.89%
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
All of the above (hope that helps!)