Answer:
0.4167 ft/s
Explanation:
The law of conservation is applied to point A and B.
This gives:
Hence, 
is the kinetic energy at the position A
is the velocity at point A
Considering point A, the kinetic energy at the point will be:
The potential energy will be:
Hence, 
is the potential energy and 
is the kinetic energy
The potential energy is given by the following:
substituting 1 lb for mg gives 0.4167 ft for r
Then the velocity, 
Answer:
staples
Door or window
Explanation:
staples.- can hit the internal core and risk of electrocution.
Door or window - can damage the outer core
Answer: For non-precision approaches, the maximum acceptable descent rate acceptable should be one that ensures the aircraft reaches the minimum descent altitude at a distance from the threshold that allows landing in the touch down zone. Otherwise, a decent rate greater than 1000fpm is unacceptable.
Explanation: For non-precision approaches, a descent rate should be used that ensures the aircraft reaches the minimum decent altitude at a distance from the threshold that allows landing in the touchdown zone (TDZ) . On many instrument approach procedures, this distance is annotated by a visual descent point (VDP) If no VDP is annotated, calculate a normal descent point to the TDZ. To determine the required rate of descent, subtract the TDZ elevation (TDZE) from the final approach fix (FAF) altitude and divide this by the time inbound. For illustration, if the FAF altitude is 1,000 feet mean sea level (MSL), the TDZE is 200 feet MSL and the time inbound is two minutes, an 400 fpm rate of descent should be applied.
A descent rate greater than approximately 1,000 fpm is unacceptable during the final stages of an approach (below 1,000 feet AGL). Operational experience and research shows that this is largely due to a human perceptual limitation that is independent of the airplane or helicopter type. As a result, operational practices and techniques must ensure that descent rates greater than 1,000 fpm are not permitted in either the instrument or visual portions of an approach and landing operation.
Answer:
rate of flow of energy(in)=14.3514 KW
Rate of flow of energy(out)=17.8092 KW
change of energy flow=3.4578 KW
Explanation:
The explanation is given in the attachments.
