Answer:
if the friction was removed from the ramp, the acceleration would drasticly increase, making it hard to stop.
Explanation:
Answer:
During start total turns
![N_1 = 296 turn](https://tex.z-dn.net/?f=N_1%20%3D%20296%20turn)
After half of the time total turns
![N_3 = \frac{29.6 + 0}{2}(5) = 74 turns](https://tex.z-dn.net/?f=N_3%20%3D%20%5Cfrac%7B29.6%20%2B%200%7D%7B2%7D%285%29%20%3D%2074%20turns)
Total number of turns during it stop
![N_2 = \frac{59.17 + 0}{2}(31) = 917 turn](https://tex.z-dn.net/?f=N_2%20%3D%20%5Cfrac%7B59.17%20%2B%200%7D%7B2%7D%2831%29%20%3D%20917%20turn)
After half of the time total turns
![N_4 = 688 turns](https://tex.z-dn.net/?f=N_4%20%3D%20688%20turns)
Explanation:
Initially the machine is at rest and then starts rotating with speed 3550 rev/min
now we will have
![f = \frac{3550}{60} = 59.17 rev/s](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B3550%7D%7B60%7D%20%3D%2059.17%20rev%2Fs)
now we know that it took 10 s to reach the speed
so angular acceleration is given as
![\alpha = \frac{\omega_f - \omega_i}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5Comega_f%20-%20%5Comega_i%7D%7Bt%7D)
![\alpha = \frac{59.17- 0}{10} = 5.917 rev/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B59.17-%200%7D%7B10%7D%20%3D%205.917%20rev%2Fs%5E2)
now it stops in 31 s so the angular deceleration is given as
![\alpha_2 = -\frac{59.17}{31} = -1.91 rev/s^2](https://tex.z-dn.net/?f=%5Calpha_2%20%3D%20-%5Cfrac%7B59.17%7D%7B31%7D%20%3D%20-1.91%20rev%2Fs%5E2)
now initially number of turn to reach the given speed
![N_1 = \frac{59.17 + 0}{2}(10) = 296 turn](https://tex.z-dn.net/?f=N_1%20%3D%20%5Cfrac%7B59.17%20%2B%200%7D%7B2%7D%2810%29%20%3D%20296%20turn)
number of turns during it stop
![N_2 = \frac{59.17 + 0}{2}(31) = 917 turn](https://tex.z-dn.net/?f=N_2%20%3D%20%5Cfrac%7B59.17%20%2B%200%7D%7B2%7D%2831%29%20%3D%20917%20turn)
Now during startup speed after t = 5 s is given as
![\omega_1 = (5.917)(5) = 29.6 rev/s](https://tex.z-dn.net/?f=%5Comega_1%20%3D%20%285.917%29%285%29%20%3D%2029.6%20rev%2Fs)
![N_3 = \frac{29.6 + 0}{2}(5) = 74 turns](https://tex.z-dn.net/?f=N_3%20%3D%20%5Cfrac%7B29.6%20%2B%200%7D%7B2%7D%285%29%20%3D%2074%20turns)
now during it stop the speed after half the time is given as
![\omega_2 = 59.17 - (1.91)(15.5) = 29.56 rev/s](https://tex.z-dn.net/?f=%5Comega_2%20%3D%2059.17%20-%20%281.91%29%2815.5%29%20%3D%2029.56%20rev%2Fs)
now the number of turns is given as
![N_4 = \frac{59.17 + 29.56}{2}(15.5)](https://tex.z-dn.net/?f=N_4%20%3D%20%5Cfrac%7B59.17%20%2B%2029.56%7D%7B2%7D%2815.5%29)
![N_4 = 688 turns](https://tex.z-dn.net/?f=N_4%20%3D%20688%20turns)
Answer:
<h3>A, I think but I'm sure </h3>
Explanation:
ur welcomeee ♥️♥️
No, interatomic bonds aren't broken during boiling.
Yes, intermolecular forces are overcome when water boils.
<h3><u>Explanation:</u></h3>
Boiling of water is a physical change of state that converts the liquid water into water vapour. This process takes place in 100°C.
In liquid water, there are different forces of attraction between the molecules of water, most importantly the van der Waal's force, and the hydrogen bonds. van der Waal's forces are very weak, so of negligible importance. But hydrogen bonding are of significant strength. They are present between the hydrogen atoms and the oxygen atoms of different molecules of water. During boiling, these forces are weakened by the increasing kinetic energy of molecules and the water molecules start to boil off as water vapour. There's no breaking of interatomic bonds, which would have produced hydrogen and oxygen gases instead of water vapour.