Viscosity is a fluids resistance to __________. a. flow c. liquify b. freeze d. vaporize

How could the movement of tectonic plates create another supercontinent like Pangaea

Margaret [11]

When the landmasses are again collected together in a specific area

The Pangea theory supports the theory of plate tectonics because it is based on the idea that all the present continents were one super-continent. For reasons still unknown, the super-continent got separated and created separate smaller continents. This theory also supports the fact that this movement will continue until the day arrives when all the smaller continents will again merge in a different configuration to create another super-continent.

4 0

3 years ago

2. Dominant trait: cleft chin (C) Mother’s gametes: Cc

andre [41]

.2. Offspring Genotypes will be Cc or cc.

Offspring phenotypes : Cleft chin or no cleft chin.

% chance child will have cleft chin: 50%

3. % chance child will have arched feet: 25%

4. % chance child will have blonde hair: 50%

5. % chance child will have normal vision: 25%

Explanation:

CASE 1 :

Dominant trait: cleft chin (C)

Recessive trait: lacks cleft chin (c)

Father’s gametes: cc

Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents= $\frac{1}{2}C + \frac{1}{2} c$ ........(i)

Gametes of cc parents = $\frac{1}{2}c + \frac{1}{2}c$ .........(ii)

Combining (i) and (ii) we get,

$\frac{1}{2} Cc + \frac{1}{2} cc$

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin = $\frac{0.5}{1}$ ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

A from mother, a from father =Aa

a from mother, A from Father = Aa

a from mother, a from father = aa

Gametes of Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

Gametes of other Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

..................................................................................

$\frac{1}{4} AA + \frac{1}{4} Aa$

+ $\frac{1}{4} Aa$ + $\frac{1}{4} aa$

..........................................................................................

$\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa$

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = $\frac{0.25}{1}$ × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive (bb)

Father’s gametes: Heterozygous (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in half Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous (Ff)

Father’s gametes: Heterozygous (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF then phenotype farsightedness

Genotype Ff then phenotype farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

3 0

3 years ago

A normal fault is formed from compressional stress. select one: a. true b. false

Dmitry_Shevchenko [17]

Answer: False.

Explanation:

A normal fault is a type of fault in which the hanging wall drop down in relative to the footwall. It is caused by tensional stress, when two block of rocks pull apart from each other. Normal fault normal happen in areas that are undergoing extension or stretching.

Reverse fault are caused by compressive stress.

8 0

3 years ago

Which connective tissue layer directly surrounds each axon in a nerve?

gregori [183]

Answer:

Endonerium

Explanation:

The endoneurium (also called endoneurial channel, endoneurial sheath, endoneurial tube, or Henle's sheath) is a layer of delicate connective tissue around the myelin sheath of each myelinated nerve fiber in the peripheral nervous system. Its component cells are called endoneurial cells.The endoneuria with their enclosed nerve fibers are bundled into groups called nerve fascicles, each fascicle within its own protective sheath called a perineurium. In sufficiently large nerves multiple fascicles, each with its blood supply and fatty tissue, may be bundled within yet another sheath, the epineurium.

4 0

3 years ago

Read 2 more answers

Sound quality can be divided

iris [78.8K]

Sound quality can be divided into amplitude, timbre and pitch. If there’s an impedance mismatch between your two devices connected to the single output, you could have a large mismatch between the levels arriving at each device. If the difference is large enough, one device may have distorted or inaudible audio.

To avoid this, you should ensure that both devices connected to the split signal are similar - such as 2 pairs of headphones, 2 recorder inputs, and so on. When you place 2 devices with wildly differing load impedances on a splitter is when you’ll encounter problems - such as headphones on one split and a guitar amp input on the other.

To get around this, you can use either a distribution amplifier (D.A.) or a transformer balanced/isolated splitter - which will work over a larger range of load impedances, typically. Depends on the quality of the splitter and the exact signal path. If you’re using the splitter to hook two things into one input, and you’re using quality connectors, you probably won’t lose much quality. There can be an increase in impedance of the cable due to the imperfect continuity of the physical connection, however with unbalanced line-level signals, impedance at both ends of the chain tends to be orders of magnitude higher than the connection will create, so one split will be barely noticeable. So too, the noise increase from the additional length of cable.

Now, one source into two inputs, that will by basic math and physics result in a 3dB drop in signal strength, which will reduce SNR by about that much. By splitting the signal path between two inputs of equal impedance, half of the wattage is being consumed by one input and half by the other (the equation changes if the inputs have significantly different impedances). So each input gets half the wattage produced by the source to drive the signal on the input cable, and in decibel terms a halving of power is a 3dB reduction. Significant, until you just turn the gain back up. The “noise floor” will be raised by however much noise is inherent in the signal path between the split and the output of the gain stage; for pro audio this is usually infinitesimal, but consumer audio can have some really noisy electronics, both for lower cost and because you’re not expected to be “re-amping” signals several times between the source and output.

8 0

3 years ago