Answer:
Explanation:
given,
charge of two spherical drop = 0.1 nC
potential at the surface = 300 V
two drops merge to form a single drop
potential at the surface of new drop = ?
r = 0.003 m
volume = 
= 
= 2.612 × 10⁻⁷ m³
![R =\sqrt[3]{\dfrac{2.612 \times 10^{-7}\times 3}{4\times \pi}}](https://tex.z-dn.net/?f=R%20%3D%5Csqrt%5B3%5D%7B%5Cdfrac%7B2.612%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%203%7D%7B4%5Ctimes%20%5Cpi%7D%7D)
R = 0.00396 m
The downwards component of weight of the object would be 35sin25 degrees = 14.8
F=14.8 - 8 = 6.8 N m= 35/9.8= 3.57 kg
F=mA
Therefore,
3.57A= 6.8 N
=> A= 6.8/3.57
=> A= 1.902 ms^-2
F(max)= (U)R where (U)= coefficient of friction
and R= Normal reaction force
R= 35cos25
= 31.72 N
Since, F(max)= 8
8= (U)* 31.72
=>(U)= 8/31.72
=>(U)= 0.25
Answer:
p = 20 kg•m/s
KE = 100 J
Explanation:
In an elastic collision of identical masses, the two masses will exchange momentums. Therefore Block 1 initially moving at 10 m/s will be moving at 2 m/s, and Block 2 will go from 2 m/s to 10 m/s
momentum = mv = 2(10) = 20 kg•m/s
KE = ½mv² = ½(2)10² = 100 J
Unfortunately, your answer selection does not have this answer as an option.
I don’t know the answer i just need the points
There are many factors which contributes as to how a machine will be processing the input energy and convert it to output energy. Even with identical mechanism, these factors will have major effect on the output. Some factors are deflection, friction and wear. Some system maybe exposed to poor lubrication than the other which'll produce more friction and wear thus lower mechanical advantage.
