Answer : The volume of water in graduated cylinder is 15.5 mL.
Explanation :
As we know that for the measurement of the volume of liquid in graduated cylinder are shown by placing the graduated cylinder on the flat surface and then view the height of the liquid in the graduated cylinder with the naked eyes directly level with the liquid.
The liquid will tend to curve downward that means this curve is known as the meniscus.
In the case of colored liquid, we are always read the upper meniscus of the liquid for the measurement.
In the case of colorless liquid, we are always read the lower meniscus of the liquid for the measurement.
In the given image, there are 5 larger and 5 smaller division between the 15 and 20 and the solution is colored. The value of larger division is 1 mL and smaller division is 0.5 mL.
So, we will read the upper meniscus of the liquid for the measurement.
Hence, the volume of water in graduated cylinder is 15.5 mL.
The answer is d
F=ma. so by substituting we find that a=4
and a=vi-vf/t
by substiting we find vf=8
so the difference is vf-vi = 8m/s
Gas is a state of matter that has no fixed shape and no fixed volume. Gas particles spread out and are evenly spaced throughout a container. ... Gas particles spread out to fill a container evenly, unlike solids and liquids
Answer: C = Q/4πR
Explanation:
Volume(V) of a sphere = 4πr^3
Charge within a small volume 'dV' is given by:
dq = ρ(r)dV
ρ(r) = C/r^2
Volume(V) of a sphere = 4/3(πr^3)
dV/dr = (4/3)×3πr^2
dV = 4πr^2dr
Therefore,
dq = ρ(r)dV ; dq =ρ(r)4πr^2dr
dq = C/r^2[4πr^2dr]
dq = 4Cπdr
FOR TOTAL CHANGE 'Q', we integrate dq
∫dq = ∫4Cπdr at r = R and r = 0
∫4Cπdr = 4Cπr
Q = 4Cπ(R - 0)
Q = 4CπR - 0
Q = 4CπR
C = Q/4πR
The value of C in terms of Q and R is [Q/4πR]
24 c 25 b 26 a 27 a 28 d 29 d i think