Answer:
The height jumped by the person on the moon is 6 times the height jumped by the person on earth.
Explanation:
As we know that the acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth.
So, it is false.
Let the mass of man is m and the gravity on moon is g' = g/6.
Let the height jumped on earth is h and the height jumped on moon is h'.
So,
m x g' x h' = m x g x h
g/6 x h' = g x h
h' = 6 h
So, the height jumped by the person is 6 times the height jumped by the person on earth.
B. is not a validated bu experimentation
120 km/h = 33.33 m/s 10.8 km/h/s = 3 m/s/s the motorist will cover a distance of 33.33 m x T(seconds) the police officer will cover a distance of 1/2 (3) T^2 they will be at the same point when 33.33T = 1.5 T^2 33.33 = 1.5 T 3T = 66.66 T = 22.22 seconds it will take the officer 22.22 seconds to catch the motorist. the officer will be moving V=AT = 3m/s x 22.22 seconds = 66.66 m/s almost 240 km/h (239.976 km/h)
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:
D try that answer if it's not correct try A
Explanation:
