Answer:
The acceleration of the centre of mass of spool A is equal to the magnitude of the acceleration of the centre of mass of spool B.
Explanation:
From the image attached, the description from the complete question shows that the two spools are of equal masses (same weight due to same acceleration due to gravity), have the same inextensible wire with negligible mass is attached to both of them over a frictionless pulley; meaning that the tension in the wire is the same on both ends.
And for the acceleration of both spools, we mention the net force.
The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.
For this system, the net force on either spool is exactly the same in magnitude because the net force is a difference between the only two forces acting on the spools; the tension in the wire and their similar respective weights.
With the net force and mass, for each spool equal, from
ΣF = ma, we get that a = ΣF/m
Meaning that the acceleration of the identical spools is equal also.
Hope this Helps!
Well, actually a heat engine using the Carnot cycle whose cold reservoir is maintained at 0 K could in principle be 100% efficient, even if the hot reservoir was merely at the ambient temperature. Since it costs no energy at all to maintain a heat reservoir at the ambient temperature as long as you have a perfect thermal connection to a large thermal capacitor like your swimming pool or a nearby aquifer you’d have built a perpetual motion machine.
Answer:
Product, ![P=9.1321972\times 10^7\times 7.004\times 10^{-3}](https://tex.z-dn.net/?f=P%3D9.1321972%5Ctimes%2010%5E7%5Ctimes%207.004%5Ctimes%2010%5E%7B-3%7D)
P = 639619.091888
Explanation:
In this case, we need to find the product of two numbers. These are as follows :
![N_1=91321972=9.13\times 10^7](https://tex.z-dn.net/?f=N_1%3D91321972%3D9.13%5Ctimes%2010%5E7)
![N_2=0.007004=7.004\times 10^{-3}](https://tex.z-dn.net/?f=N_2%3D0.007004%3D7.004%5Ctimes%2010%5E%7B-3%7D)
On doing calculation, we get the product of two numbers as :
![P=N_1\times N_2](https://tex.z-dn.net/?f=P%3DN_1%5Ctimes%20N_2)
![P=9.1321972\times 10^7\times 7.004\times 10^{-3}](https://tex.z-dn.net/?f=P%3D9.1321972%5Ctimes%2010%5E7%5Ctimes%207.004%5Ctimes%2010%5E%7B-3%7D)
P = 639619.091888
In scientific notation the product of two number is given by :
![P=63\times 10^4](https://tex.z-dn.net/?f=P%3D63%5Ctimes%2010%5E4)
Hence, this is the required solution.