Explanation:
(a) As it is given that there is no vertical displacement. Hence, the forces on y-axis are equal and opposite.
Also, it is known that F = mg
So, F =
= 548.8 N
Also,
where, = coefficient of kinetic friction
N = force calculated
Putting the values into the formula as follows.
=
= 115.25 N
Hence, the magnitude of the frictional force is 115.25 N.
(b) Now, let us assume that the acceleration is towards the positive x-direction.
a = 7.74
Therefore, the magnitude of the crate's acceleration is 7.74 .
sample was gray, non-metallic, had
density of 3.8 g/mL, and a volume
of 48.3 mL
Of the observed properties listed
below, which is considered to be an
extensive property?
A Gray
B Metallic
C Density
D Volume
Answer:
D. Volume
Explanation:
Properties mentioned in the question;
colour = gray
Nonmetallic
Density = 3.8 g/mL
Volume = 48.3 mL
Of the mentioned, which is extensive ?
Extensive properties are those properties that are proportional to the amount of substance present.
The correct option is D. Volume. The other properties do not change with increasing or decreasing amount of the mineral
Volume is considered to be an extensive property.
The extensive property out of the observed properties listed is volume. An extensive property is one that depends on the amount of a substance. In this case, the volume of the mineral sample is directly proportional to the amount of the sample, so it is considered an extensive property.
Learn more about Extensive property here:
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B. A sample of 2.00 g of O2 is removed.
C. A sample of 4.00 g of O2 is added to the 4.80 g of O2 gas in the container.
By using the ideal gas law and molar mass calculations, the final volumes are found to be A. 65.0 L, B. 8.75 L, and C. 27.5 L.
To calculate the final volume when additional O2 is added or when some O2 is removed, we can use the concept of the molar mass and the ideal gas law that states that volume is directly proportional to the amount of gas, assuming pressure and temperature is constant.
The molar mass of O2 is approximately 32.00 g/mol.
A. 0.500 moles of O2 is added. This equals 0.500 * 32 g = 16 g. The total mass in the system is now 20.8 g. If the original 15.0 L represented 4.80 g, now 20.8 g would represent 15.0 L * 20.8/4.80 = 65.0 L.
B. 2.00 g of O2 is removed. So, the total mass in the system is now 2.80 g. If the original 15.0 L represented 4.80 g, now 2.80 g would represent 15.0 L * 2.80/4.80 = 8.75 L.
C. 4.00 g of O2 is added. So, the total mass in the system is now 8.80 g. If the original 15.0 L represented 4.80 g, now 8.80 g would represent 15.0 L * 8.80/4.80 = 27.5 L.
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b. q = Hvm
c. P₁V₁ = P₂V₂
d. D = m/V
e. K = C + 273
H2(g) = 2.0 moles per liter
H2O(g) = 4.0 moles per liter
CO2(g) = 5.0 moles per liter
CO(g) = 4.0 moles per liter
What is the equilibrium constant, Keq, for this reaction?