Answer:
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.
determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.
a)0.7956kg/s
b)5.437 × 10⁻³m²
Explanation:
The concepts related to the change of mass flow for both entry and exit is applied
The general formula is defined by
Where,
values are divided by inlet(1) and outlet(2) by
PART A) Applying the flow equation
PART B) For the exit area we need to arrange the equation in function of Area, that is
Answer:
The total work that the rope does to Mangnus is - 5780 Jules.
Explanation:
By definition, the work is defined as:
Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.
Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.
For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement:
Answer:
Explanation:
3.75 * 10^-7
=3.75 * 1/10^7
=3.75/10000000
=3/800000000
any base which has it's power negative do it's reciprocal then the power will be positive.
Answer:
So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3
