Answer:
19.3
Explanation:
Assuming we have to find Specific gravity of gold.
As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object
so it is given by
specific gravity = weight of the object/weight of the water displaced
now we have
weight of the object = (density)(volume)g
weight of object = (19.3)(0.55)g
now weight of the liquid displaced is given by
weight of water displaced = (1 g/cm^3)(0.55ml)g
now we have
specific gravity = (19.3×0.55)/(1×0.55)
specific gravity= 19.3
Answer:
v =
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:


Looks like you simply substitute the length of the femur
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
B) atmosphere pressure, i believe