Answer:
The child represented by a star on the outside path.
Explanation:
Solution :
Speed of the air craft, = 262 m/s
Fuel burns at the rate of, = 3.92 kg/s
Rate at which the engine takes in air, = 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft, =921 m/s
Therefore, the upward thrust of the jet engine is given by
F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
=
Therefore thrust of the jet engine is .
Answer:
a)
b)
c)
d) Displacement = 22 m
e) Average speed = 11 m/s
Explanation:
a)
Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:
Therefore, acceleration is
b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:
c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:
d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):
Displacement =
e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:
Average velocity =
Answer:
troposphere and stratosphere
Seriously this is fun dont know why u want me to enjoy this so ummmm hmmm lolololol the equation for this would be s=vif+1/2at^2 so time is lets see plug it and vell vi=0 since it started from rest so s=1/2at^2 and plug it in the displacement and time 110=1/2 a 5.21^2 and solve it and u get 8.1m/s^2