The answers are solar for the first one, and chemical for the second one
Answer:
The correct answer is - Both move with constant speed.
Explanation:
If an object moves in a particular way that it covers the same distance at the same times the speed considered as constant suppose if two objects a and b both travel 10 m and 20 m respectively and the time they take in travel are 2 and 4 seconds respectively than we can say they both have a constant speed.
The relation between the variables x,t, and v is
x=v.t
If V is constant which is the speed we get a straight line x-t in the graph.
By the given information we can say that both objects A and B are traveling at a constant speed.
..............4 is D and 6 is A
Answer:
Scientists often use models of Earth to study the movement of earth and it's moon. They figure out what the distance is from the moon to the point between earth and moon where the gravitational pulls of earth and moon are equal.
Answer/ Explanation:
a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY
The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.
b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr
, Vg+Y
, VgXr
, VgY
. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.
c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye
