QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.
ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
Answer:
4.2 km
Explanation:
A triangle is a polygon with three sides and three angles. There are different types of triangles such as isosceles triangle, scalene triangle, equilateral triangle and right angled triangle.
Sine rule states that for a triangle with angles A,B and C and corresponding opposite sides a, b, and c, the following formula holds:
Let the starting point be A, hence the side opposite to the angle = a = distance travelled in north east direction.
Let the point of 60° of northeast be B = 90° + (90° - 60°) = 120°, hence the side opposite to the angle = b = resultant displacement = 8 km.
Let C be the endpoint, hence the side opposite to the angle = c = distance travelled north = 5 km
Using sine rule:
∠A + ∠B + ∠C = 180° (sum of angle in a triangle)
∠A + 120 + 32.8 = 180
∠A = 27.2°
Therefore she travelled 4.2 km in the north east direction
Answer:
0.99kgm/s
Explanation:
Given parameters:
Initial momentum = 5.8kgm/s
Final momentum = 6.79kgm/s
Unknown:
Magnitude of the impulse applied to the baseball = ?
Solution:
To solve this problem;
The impulse applied to the baseball can be found by using the expression below:
Impulse = Final momentum - Initial momentum
Impulse = 6.79kgm/s - 5.8kgm/s = 0.99kgm/s
Answer:
The
Explanation:
The horizontal distance covered is known as the range expressed as;
R = U√2H/g
U is the speed = 200m/s
H is the max height = 800m
g is the gravity = 9.8m/s²
R = 200√2(800)/9.8
R = 200√1600/9.8
R = 200√163.27
R = 200(12.78)
R = 2555.54
Hence the package should be released at a distance of 2555.54m
