Answer:

Explanation:
Given data
Speed of jet Vjet=1190 km/h
Speed of prop driven Vprop=595 km/h
Height of jet 7.5 km
Height of prop driven transport 3.8 km
Density of Air at height 10 km p7.8=0.53 kg/m³
Density of air at height 3.8 km p3.8=0.74 kg/m³
The drag force is given by:

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:

Answer:
Equal to the weight of the object
Explanation:
There are two forces acting on the object hanging on the string:
- The tension in the string, T, acting upward
- The weight of the object, mg, acting downward
We can write Newton's second law as:

where a is the acceleration of the object.
Here we are told that the elevator is moving at constant speed: this means that the acceleration of the object is zero, therefore a = 0 and the equation becomes

therefore, the tension in the string is equal to the weight of the object.
New sound intensity level after the move is
10 log (1/3²)
= -20 log (3)
= -20 (0.4771) = 9.54 dB LOWER than before the move.
The force and amount of attracton is great for larger masses