Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. spec
ifically identify each quantity in your example.
1 answer:
The example I've given is shown as an illustration in the picture. Suppose a jogger covers 2 km as he heads to the east. Then, he heads up north for another 1.5 km. Now, if you are to find the distance, you simply have to take the sum.
Distance = 2 + 1.5 = 3.5 km
If you are to find the displacement, you find the distance from the starting point to the end point, represented by the red arrow. Using the pythagorean theorem,
Displacement = √(2 km)² + (1.5 km)² = 2.5 km
If you are to find the magnitude of displacement, this is equivalent to distance. So, it is equal to 3.5 km. The displacement is a vector quantity, which presents the magnitude and the direction. The magnitude of the displacement does not take into account the direction of motion, just the total distance covered.
Distance = 2 + 1.5 = 3.5 km
If you are to find the displacement, you find the distance from the starting point to the end point, represented by the red arrow. Using the pythagorean theorem,
Displacement = √(2 km)² + (1.5 km)² = 2.5 km
If you are to find the magnitude of displacement, this is equivalent to distance. So, it is equal to 3.5 km. The displacement is a vector quantity, which presents the magnitude and the direction. The magnitude of the displacement does not take into account the direction of motion, just the total distance covered.
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Answer:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be
Let the exit pressure be
Ratio of reservoir pressure and exit pressure
= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s