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
![$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$](https://tex.z-dn.net/?f=%24F%3DS_%7Bair%7D%28S_%7Bexh%7D-S_a%29%2B%28S_b%20%5Ctimes%20S_%7Bexh%7D%29%24)
F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
= ![$4.8 \times 10^6 \ N$](https://tex.z-dn.net/?f=%244.8%20%5Ctimes%2010%5E6%20%5C%20N%24)
Therefore thrust of the jet engine is
.
Force = (mass) x (acceleration)
= (2 kg) x (20 m/s²) = 40 newtons .
about 9 pounds .
Note: The ball can only accelerate during the short time
that the clubface is in contact with it. Once it leaves the
clubface, it can't accelerate any more.
Also ... that's one heckuva golf ball. It weighs about 4.4 pounds !
Answer:
10.25 seconds
Explanation:
Divide 400/39, leaves you with how many seconds it takes to travel that distance.
It's nice that this problem has everything in meters already, so that you don't have to do any unit conversions.
Answer:
Angular momentum, ![L=6.47\times 10^{-3}\ m](https://tex.z-dn.net/?f=L%3D6.47%5Ctimes%2010%5E%7B-3%7D%5C%20m)
Explanation:
It is given that,
Radius of the axle, ![r=3.21\ mm=3.21\times 10^{-3}\ m](https://tex.z-dn.net/?f=r%3D3.21%5C%20mm%3D3.21%5Ctimes%2010%5E%7B-3%7D%5C%20m)
Tension acting on the top, T = 3.15 N
Time taken by the string to unwind, t = 0.32 s
We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :
![\tau=\dfrac{dL}{dt}](https://tex.z-dn.net/?f=%5Ctau%3D%5Cdfrac%7BdL%7D%7Bdt%7D)
Torque acting on the top is given by :
![\tau=F\times r](https://tex.z-dn.net/?f=%5Ctau%3DF%5Ctimes%20r)
Here, F is the tension acting on it. Torque acting on the top is given by :
![\tau=2F\times r](https://tex.z-dn.net/?f=%5Ctau%3D2F%5Ctimes%20r)
![2T\times r=\dfrac{L}{t}](https://tex.z-dn.net/?f=2T%5Ctimes%20r%3D%5Cdfrac%7BL%7D%7Bt%7D)
![L=2T\times r \times t](https://tex.z-dn.net/?f=L%3D2T%5Ctimes%20r%20%5Ctimes%20t)
![L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32](https://tex.z-dn.net/?f=L%3D2%5Ctimes%203.15%5Ctimes%203.21%5Ctimes%2010%5E%7B-3%7D%5Ctimes%200.32)
![L=6.47\times 10^{-3}\ m](https://tex.z-dn.net/?f=L%3D6.47%5Ctimes%2010%5E%7B-3%7D%5C%20m)
So, the angular momentum acquired by the top is
. Hence, this is the required solution.