Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o
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Answer
The frequency and the period are 45 cycles/second and 0.023 seconds
Explanation:
The frequency of the engine is given by :
We have, the engine increases its performance from zero to 2700 rpms.
Frequency,
The time period of the engine is given by :
T = f/f
So,
Hence, the required frequency and the period are 45 cycles/second and 0.023 seconds respectively.
Answer:
See explanation
Explanation:
Solution:-
- We will first set a datum as the free surface of water in the pool.
- There is a square window with side length ( L = 8 m ) on the vertical side of the pool.
- The depth of the pool is given to be 54 m. The top of the window is ( d = 43 m ) from the free surface.
- We are to determine the resultant hydro static force acting on the window.
- Hydro static force ( F ) acting on an object of area ( A ) fully immersed in fluid is given by the following relationship as follows:
F = Pc*A
Where,
Pc: The hydrostaic pressure acting on the centroid of the obect.
- The hydrostatic pressure acting on the centroid of the object immersed in any fluid can be expressed by the following defining relationship:
Pc = γ*yc
Where,
γ: The specific weight of the fluid
yc: The vertical distance from free-surface to the centroid.
- Assuming homogeneous distribution of material used for the window of square shape. The centroid coincides with the geometric center of the window which is as a distance ( yc ) from the free-surface:
- Now we can calculate the resultant hydro static force ( F ) acting on the window. The specific gravity of fluid ( water ) is γ = 9.8KN/m^3.
Note: The values given in the posted question seem unreasonable. I have assumed values in order of convenience. However, the procedure of solving the problem remains exactly the same.