400 mA, 70 ms is the following exposure factors will produce the greatest receptor exposure.
C: 400 mA, 70 ms
<u>Explanation:</u>
As SID builds, the introduction rate diminishes and receptor presentation diminishes. SID and the mAs required to keep up the introduction to the IR have a straightforwardly corresponding relationship (as the SID builds, the mAs required to keep up presentation to the IR increments by a corresponding sum).
An expanded SID likewise decreases amplification (size twisting). The most extreme SID ought to be utilized when conceivable to limit amplification. Infrequently, however, the SID can be purposefully diminished for amplification. SID influences size yet not shape twisting.
Answer:
squamous epithelium consists of a single layer of flat cells. It is found in lungs,heart and blood vessels.It allows the movement of materials across it.
stratified epithelium has many layer of flat cells.It is present in lining of oesophagus and mouth and also over the skin
Answer:
25% or 1/4
Explanation:
The gene for colour in Heliodors is controlled by two contrasting alleles that codes for Red (R) and Yellow (Y) colours. However, these two alleles exhibit incomplete dominance, which is a phenomenon whereby a combination of both alleles gives rise to a third intermediate phenotype that is a blending of the other two parental phenotypes. In this case, both colours gives rise to a heterozygous Orange coloration (RY) in Heliodors.
However, if two orange Heliodors (RY) are crossed, four possible offsprings will be produced with the genotypes: RR, RY, RY, YY. This shows a phenotypic ratio of 1 red: 2orange: 1yellow. Hence, the probability of having a child with red coloration is 1 out of 4 possible offsprings i.e. 1/4.
Expressing this in percentage, we have 1/4 × 100 = 25%.
Answer: Stomata
Explanation:
Stomata are pores in the leaf that allow gas exchange where water vapor leaves the plant and carbon dioxide enters.
You can gather information from both actions.
