It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
![F_{r}=kv^2](https://tex.z-dn.net/?f=F_%7Br%7D%3Dkv%5E2)
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
![F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}](https://tex.z-dn.net/?f=F%5Csin%5Ctheta-F_%7Br%7D%3Dm%5Cdfrac%7Bd%5E2v%7D%7Bdt%5E2%7D)
Put the value into the formula
![3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}](https://tex.z-dn.net/?f=3200%5Csin%2030-kv%5E2%3Dm%5Cdfrac%7Bd%5E2v%7D%7Bdt%5E2%7D)
Put the value of k
![3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}](https://tex.z-dn.net/?f=3200%5Ctimes%5Cdfrac%7B1%7D%7B2%7D-v%5E2%3Dm%5Cdfrac%7Bd%5E2v%7D%7Bdt%5E2%7D)
![1600-v^2=m\dfrac{d^2v}{dt^2}](https://tex.z-dn.net/?f=1600-v%5E2%3Dm%5Cdfrac%7Bd%5E2v%7D%7Bdt%5E2%7D)
At terminal velocity ![\dfrac{d^2v}{dt^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2v%7D%7Bdt%5E2%7D%3D0)
So, ![1600-v^2=0](https://tex.z-dn.net/?f=1600-v%5E2%3D0)
![v=\sqrt{1600}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B1600%7D)
![v=40\ ft/sec](https://tex.z-dn.net/?f=v%3D40%5C%20ft%2Fsec)
Hence, The velocity is 40 ft/sec.
Answer:
c. 981 watts
![P=981\ W](https://tex.z-dn.net/?f=P%3D981%5C%20W)
Explanation:
Given:
- horizontal speed of treadmill,
![v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}](https://tex.z-dn.net/?f=v%3D100%5C%20m.min%5E%7B-1%7D%3D%5Cfrac%7B5%7D%7B3%7D%20%5C%20m.s%5E%7B-1%7D)
- weight carried,
![w=588.6\ N](https://tex.z-dn.net/?f=w%3D588.6%5C%20N)
- grade of the treadmill,
![G\%=10\%](https://tex.z-dn.net/?f=G%5C%25%3D10%5C%25)
<u>Now the power can be given by:</u>
![P=v.w](https://tex.z-dn.net/?f=P%3Dv.w)
(where grade is the rise of the front edge per 100 m of the horizontal length)
![P=981\ W](https://tex.z-dn.net/?f=P%3D981%5C%20W)
Answer:
Sea-floor spreading occurs in the oceanic ridges. In there, volcanic activity, together with the gradual movement of the bottom, form new oceanic crust. This allows a better understanding of the continental drift explained by the theory of plate tectonics.
The greatest evidence for Sea-floor spreading is the oceanic trenches, the oceanic ridges, the magma protruding to the surface and the new seafloor.
In previous theories, continents were assumed to be transported across the sea. Harry Hess, in the 1960s, proposed the idea that the seabed itself moves as it expands from a central point. The theory is now accepted, and the phenomenon is thought to be caused by convection currents in the upper layer of the mantle.
Answer:
atm
Explanation:
The pressure at the bottom of any liquid column is equal to product of density of the liquid , gravitational acceleration constant (g) and height of the water column
Thus, ![P = \rho*g*h](https://tex.z-dn.net/?f=P%20%3D%20%5Crho%2Ag%2Ah)
Substituting the given values, we get -
kg/m3
m/s^2
meters
atm