A wind tunnel is a traditional tool used for trans-sonic flight research. But, during those days, it’s not exact and consistent to use for aircraft flow conditions. The reason is that wind tunnel models will be visible to airflow that produces shock waves. So the U.S and Britain develop an instrument to get data and record flow conditions for a specified speed, these are the specialized research airplanes.

6 0

3 years ago

Hellppppppppppp please thanks

tester [92]

Point C would the greatest

8 0

3 years ago

A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires

kykrilka [37]

The answer is 8.8 m/s^2. The acceleration (a) is the change in velocity (v) in time (t): a=Δv/t. Δv=v2-v1. Since the car starts from the rest, v1=0 m/s. The final velocity (v2) can be calculated from kinetic energy (KE) and the mass (m) of the car since KE=1/2*m*v2^2. From here: v2^2=2KE/m. v2=√(2KE/m). KE=3.13×10^5 J=3.13×10^5 kg*m^2/s^2, m=290 kg. v2=√(2*3.13×10^5 kg*m^2/s^2/290 kg)=√(2,158.62 m^2/s^2)=46.5 m/s. So, Δv=v2-v1=46.5 m/s - 0 m/s= 46.5 m/s. Now, we have Δv = 46.5 m/s and t=5.3 s, so the acceleration is: a=Δv/t=46.5 m/s/5.3 s = 8.8 m/s^2.

6 0

4 years ago

A fireworks shell is accelerated from rest to a velocity of 56.0 m/s over a distance of 0.270 m.

deff fn [24]

Answer:

(a) The acceleration lasts 9.642 seconds.

(b) The acceleration of the fireworks shell is 5.808 meters per square second.

Explanation:

Statement is incomplete. Complete description is presented below:

A fireworks shell is accelerated from rest to a velocity of 56 meters per second over a distance of 0.270 meters. (a) How long (in seconds) did the acceleration last? (b) Calculate the acceleration.

(b) Let suppose that the fireworks shell is accelerated uniformly. Given that initial ( $v_{o}$ ) and final speed ( $v_{f}$ ), measured in meters per second, and distance ( $s$ ), measured in meters, are known, we calculate the acceleration ( $a$ ), measured in meters per square second, by this kinematic expression: