Which of the following four circuit diagrams best represents the experiment described in this problem?

Physics

1 answer:

Valentin [98]3 years ago

6 0

We don't see any circuit diagrams.

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

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During a softball game, a shortstop catch a ground ball. The action forces is the ball pushing on the glove. What is the reactio

marusya05 [52]

<span>The reaction force would be the exact opposite of the action. In this case, choice (c) would be the most correct. If the action is the ball pushing the glove, the reaction would then be the glove pushing back on the ball.</span>

4 0

3 years ago

ne true-breeding line of mice is obese and dark, and the other true-breeding line is lean and light. Dark is dominant to light,

LenKa [72]

Answer:

$\frac{3}{8}$

Explanation:

Let the allele for dark color be represented by "D" and the allele for light color be represented by "d"

Also D is dominant over d

Similarly, let the allele for obese trait in mice be represented by "O" and the allele for lean trait in mice be represented by "L"

Obese and lean exhibit incomplete dominance

Genotype of true breeding dark and obese mice be DDOO

Genotype of true-breeding lean and light mice is ddLL

Offspring from F1 cross -

DdOL

F2 cross is between DdOL * DdOL

DO DL dO dL

DO DDOO DDOL DdOO DdOL

DL DDOL DDLL DdOL DdLL

dO DdOO DdOL ddOO ddOL

dL DdOL DdLL ddOL ddLL

dark and intermediate between obese and lean offspring genotype –

2 DDOL, 4 DdOL

So total six out of sixteen offspring are dark and intermediate between obese and lean -

$\frac{6}{16} \\\frac{3}{8}$

8 0

4 years ago

A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them

Tatiana [17]

Answer:

A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

$\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}$