Answer:
The thrust of the jet engine is 4188.81 N.
Explanation:
Given that,
Speed = 260 m/s
Rate in air= 53.3 kg/s
Rate of fuel = 3.63 kg/s
Relative speed = 317 m/s
We need to calculate the rate of mass change in the rocket
Using formula of rate of mass
![\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}](https://tex.z-dn.net/?f=%5Cdfrac%7BdM%7D%7Bdt%7D%3D%5Cdfrac%7BdM_%7Ba%7D%7D%7Bdt%7D%2B%5Cdfrac%7BdM_%7Bf%7D%7D%7Bdt%7D)
Put the value into the formula
![\dfrac{dM}{dt}=53.3+3.63](https://tex.z-dn.net/?f=%5Cdfrac%7BdM%7D%7Bdt%7D%3D53.3%2B3.63)
![\dfrac{dM}{dt}=56.93\ kg/s](https://tex.z-dn.net/?f=%5Cdfrac%7BdM%7D%7Bdt%7D%3D56.93%5C%20kg%2Fs)
We need to calculate the thrust of the jet engine
Using formula of thrust
![T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7BdM%7D%7Bdt%7Du-%5Cdfrac%7BdM_%7Ba%7D%7D%7Bdt%7Dv)
Put the value into the formula
![T=56.93\times317-53.3\times260](https://tex.z-dn.net/?f=T%3D56.93%5Ctimes317-53.3%5Ctimes260)
![T=4188.81\ N](https://tex.z-dn.net/?f=T%3D4188.81%5C%20N)
Hence, The thrust of the jet engine is 4188.81 N.
The angular velocity of the cockroach does not change as long as it rotates with the disk it has the same angular velocity as the disk but by changing the distance from the center of the disk the linear velocity changes (V=r×ω) because the radius changes.
in order to find the kinetic energy you can use this formula k=1/2 × m × V²
so k/k0 = (V/V0)² = (r/r0)²
where the subscripts 0 denotes to the initial position of the cockroach so r0=2×r then we have k/k0=1/4
Voltage = Current (I) × Resistance (R)
V = 10 × 28.5 = 285v
Answer:
acceleration is the change in distance over time.
acceleration is the change in velocity over time.
u can write any one of two
hope it helps you
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