Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet
Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input = 
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
Thus, the efficiency is 85.8 %.
Answer: 1479watts
Explanation:
Power is defined as the energy expended or work done in a specific time.
Mathematically,
Power = Workdone/time taken
Since work done is force × distance
Power = force × distance/time
Force = Mary's weight = 87N
Distance = height of the flight = 102meters
Time = 6.0seconds
Substituting in the formula we have;
Power = 87 × 102/6
Power = 1,479watts
Note that the time must be in seconds before usage. If its given in minutes, you will have to convert to seconds
The average speed is given by the total distance travelled/total time taken thus the speed will be
100/50 = 2m/s
