Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.
Explanation: Hope this helped :)
Answer:
The thrust is ![6\times 10^6\ N](https://tex.z-dn.net/?f=6%5Ctimes%2010%5E6%5C%20N)
Explanation:
Given that,
Mass of gas, ![m=1.5\times 10^3\ kg](https://tex.z-dn.net/?f=m%3D1.5%5Ctimes%2010%5E3%5C%20kg)
The rate at which the gas is expelling, ![\dfrac{dv}{dt}=4\times 10^{3}\ m/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdv%7D%7Bdt%7D%3D4%5Ctimes%2010%5E%7B3%7D%5C%20m%2Fs)
We need to find the thrust produced by the gas.
We know that force is equal to the rate of change of momentum. So,
![F=\dfrac{p}{t}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bp%7D%7Bt%7D)
Also, p = mv
![F=\dfrac{mv}{t}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7Bmv%7D%7Bt%7D)
So,
![F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N](https://tex.z-dn.net/?f=F%3D1.5%5Ctimes%2010%5E3%5Ctimes%204%5Ctimes%2010%5E3%5C%5C%5C%5CF%3D6%5Ctimes%2010%5E6%5C%20N)
So, the thrust is ![6\times 10^6\ N](https://tex.z-dn.net/?f=6%5Ctimes%2010%5E6%5C%20N)
Answer:
(A) Angular speed 40 rad/sec
Rotation = 50 rad
(b) 37812.5 J
Explanation:
We have given moment of inertia of the wheel ![I=25kgm^2](https://tex.z-dn.net/?f=I%3D25kgm%5E2)
Initial angular velocity of the wheel ![\omega _0=10rad/sec](https://tex.z-dn.net/?f=%5Comega%20_0%3D10rad%2Fsec)
Angular acceleration ![\alpha =15rad/sec^2](https://tex.z-dn.net/?f=%5Calpha%20%3D15rad%2Fsec%5E2)
(a) We know that ![\omega =\omega _0+\alpha t](https://tex.z-dn.net/?f=%5Comega%20%3D%5Comega%20_0%2B%5Calpha%20t)
We have given t = 2 sec
So ![\omega =10+15\times 2=40rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D10%2B15%5Ctimes%20%202%3D40rad%2Fsec)
Now ![\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad](https://tex.z-dn.net/?f=%5CTheta%20%3D%5Comega%20_0t%2B%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2%3D10%5Ctimes%202%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2015%5Ctimes%202%5E2%3D50rad)
(b) After 3 sec ![\omega =10+15\times 3=55rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D10%2B15%5Ctimes%203%3D55rad%2Fsec)
We know that kinetic energy is given by ![Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J](https://tex.z-dn.net/?f=Ke%3D%5Cfrac%7B1%7D%7B2%7DI%5Comega%20%5E2%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2025%5Ctimes%2055%5E2%3D37812.5J)
Answer:
Nuclear Fusion
Explanation:
The process releases energy because the total mass of the resulting single nucleus is less than the mass of the two original nuclei. The leftover mass becomes energy.