The volume of the tire at the given diameter and thickness of tube is determined as 1,128.2 cubic inch.
<h3>What is volume?</h3>
Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.
<h3>
Volume of the tire</h3>
The volume of the tire is the measure of the product of area and thickness of the tire.
The volume of the tire is calculated as follows;
Radius of the tire = 0.5 x 26" = 13"
Volume of the tire = Area x thickness
Volume of the tire = πr² x h
where;
- r is the radius of the tire
- h is the thickness of the tube
Volume of the tire = π(13)² x (2.125)
Volume of the tire = 1,128.2 cubic inch
Thus, the volume of the tire at the given diameter and thickness of tube is determined as 1,128.2 cubic inch.
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Answer:
1.82°C
Explanation:
presumir inicial comprimento = l₀ = 5 m
dado que:
coeficiente de expansão (α) = 22 x 10⁻⁶, A variação no comprimento (ΔL) = 0.0002 m,
inicial comprimento = l₀ = 5 m
variação na temperatura = ΔT
para Calcular a variação na temperatura, nós usamos a fórmula:
substituindo valores:
variação na temperatura = 1.82°C
Answer:
block 2 or 4
because of the distribution of weight and force being applied to the object
This would be known to be determined by its sound wave which would be the amplitude. The pitch of sound would actually be a little differenr in this matter. But know that we know and understand the sound wave of the pitch of sound, we know that this is done by amplitude.
Answer:
Explanation:
The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:
Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.
E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.
DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.
The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.
