before adjusting drive-belt tension, technician a checks for proper pulley alignment. technician b looks up the specified belt t
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Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = ×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
The distance below the top of the cliff that the two balls cross paths is 7.53 meters.
<u>Given the following data:</u>
- Initial velocity = 0 m/s (since the ball is dropped from rest).
- Height = 30 meters.
<u>Scientific data:</u>
- Acceleration due to gravity (a) = 9.8
.
To determine how far (distance) below the top of the cliff that the two balls cross paths, we would apply the third equation of motion.
<h3>How to calculate the velocity.</h3>Mathematically, the third equation of motion is given by this formula:
<u>Where:</u>
- V is the final velocity.
- U is the initial velocity.
- a is the acceleration.
- S is the distance covered.
Substituting the parameters into the formula, we have;
V = 24.25 m/s.
<u>Note:</u> The final velocity of the first ball becomes the initial velocity of the second ball.
The time at which the two balls meet is calculated as:
Time = 1.24 seconds.
The position of the ball when it is dropped from the cliff is calculated as:
Lastly, the distance below the top of the cliff is calculated as:
Distance = 7.53 meters.
Read more on distance here: brainly.com/question/10545161