Answer:
The best option is for the following option m = 15 [g] and V = 5 [cm³]
Explanation:
We have that the density of a body is defined as the ratio of mass to volume.

where:
Ro = density = 3 [g/cm³]
Now we must determine the densities with each of the given values.
<u>For m = 7 [g] and V = 2.3 [cm³]</u>
![Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]](https://tex.z-dn.net/?f=Ro%3D7%2F2.3%5C%5CRo%3D3.04%20%5Bg%2Fcm%5E%7B3%7D%20%5D)
<u>For m = 10 [g] and V = 7 [cm³]</u>
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<u>For m = 15 [g] and V = 5 [cm³]</u>
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<u>For m = 21 [g] and V = 8 [cm³]</u>
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Answer:
In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.
Answer:
The force of gravity exerts a downward force. The floor exerts an upward force. Since these two forces are of equal magnitude and in opposite directions, they balance each other.
If the solution is treated as an ideal solution, the extent of freezing
point depression depends only on the solute concentration that can be
estimated by a simple linear relationship with the cryoscopic constant:
ΔTF = KF · m · i
ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF
(solution).
KF, the cryoscopic constant, which is dependent on the properties of the
solvent, not the solute. Note: When conducting experiments, a higher KF
value makes it easier to observe larger drops in the freezing point.
For water, KF = 1.853 K·kg/mol.[1]
m is the molality (mol solute per kg of solvent)
i is the van 't Hoff factor (number of solute particles per mol, e.g. i =
2 for NaCl).
Answer:
final velocity will be44.72m/s
Explanation:
HEIGHT=h=100m
vi=0m/s
vf=?
g=10m/s²
by using third equation of motion for bodies under gravity
2gh=(vf)²-(vi)²
evaluating the formula
2(10m/s²)(100m)=vf²-(0m/s)²
2000m²/s²=vf²
√2000m²/s²=√vf²
44.72m/s=vf