Answer:
Thus, the payback time for cavity wall insulation is 7.5 years
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law




Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula


Hence, The speed of space station floor is 49.49 m/s.
Answer:
A star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears. Therefore, if two stars have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!
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Answer:
Thrust developed = 212.3373 kN
Explanation:
Assuming the ship is stationary
<u>Determine the Thrust developed</u>
power supplied to the propeller ( Punit ) = 1900 KW
Duct distance ( diameter ; D ) = 2.6 m
first step : <em>calculate the area of the duct </em>
A = π/4 * D^2
= π/4 * ( 2.6)^2 = 5.3092 m^2
<em>next : calculate the velocity of propeller</em>
Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0
∴V^3 = Punit * 2 / A*β
= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )
hence V2 = 8.9480 m/s
<em>Finally determine the thrust developed </em>
F = Punit / V2
= (1900 * 10^3) / ( 8.9480)
= 212.3373 kN
Net work done on student B by the Ferris wheel in moving from the top to the bottom is mathematically given as
net work done on A =0.
<h3>
Net work done </h3>
Generally the equation for the work energy theorem is mathematically given as
net work done on A = change in kinetic energy of A.
Where, angular velocity is constant.
change in kinetic energy = 0.
Hence, from work energy theorem,
net work done on A =0.
For more information on work
brainly.com/question/756198