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PSYCHO15rus [73]
2 years ago
8

Scienceeee i need help asap no rocky lol please actually i need help

Biology
2 answers:
trapecia [35]2 years ago
7 0
Yeah it’s Erosion and deposition



Explanation: Flowing water picks up soil particles, and carries them along until the flow ebbs and particles sink to the bottom...Anything that reduce the energy of the flow field can cause larger particles to drop below the critical Reynolds number and drop out the fluid stream ( deposition).
zalisa [80]2 years ago
5 0

Answer: Erosion and deposition

Explanation: As the water ran through and against the river banks and land, it slowly begins to erode. Also, as deposition occurs (Deposition-When a river deposits the silt and other things in it out into the ocean or where ever it empties.) it creates a delta.

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What is the main difference between the intertidal zone and the neritic zone?
Inessa [10]

Answer:

intertidal zone is the area of shore between the high tide and low tide lines whereas neritic zone is the area of the ocean from the low tide line out to the continental shelf.

4 0
3 years ago
Read 2 more answers
2. List three sources of error that could account for the differences between your values for the enthalpy of fusion of water an
Dvinal [7]

1 trial :  nothing is given for result comparision - so we have no idea if it's a mistake.

2nd trial : The results can be compared - if varies, one may go wrong, but which one?

3rd trial : If 3rd result is different from 1st and 2nd, it is unreliable.

calculating enthalpy of fusion. M, C and m,c = mass and specific heat of calorimeter and water, n, L = mass and heat of fusion of ice; T = temperature fall.

L = (mc+MC)T/n.

c=4.18 J/gK. assuming copper calorimeter , so C=0.385 J/gK.

1. M = 409g, m = 45g. T = 22c, n = 14g

L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.

2. M = 409g, m = 49g, T = 20c, n = 13g

L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.

3. M = 409g, m = 54g, T = 20c, n = 14g

L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.

(i) Estimate error in L from spread of 3 results.

Average L = 549.3 J/g.

squared differences average (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96

standard deviation = 5.9964

standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.

% error = 3/547 x 100% = 0.5%.

(ii) Estimate error in L from accuracy of measurements:

error in masses = +/-0.5g

error in T = +/-0.5c

For Trial 3

M = 409g, error = 0.5g

m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)

n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g

K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962

L = K*T/n

% errors are

K: 3/383 x 100% = 0.77

T: 0.5/20 x 100% = 2.5

n: 1.0/14 x 100% = 7.14

% errors in K and T are << error in n, so ignore them.

% error in L = same as in n = 7% x 547.4 = 40

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.

Both are very far above  334 J/g, so there is at least one systematic error  

e.g: calorimeter may not be copper, so C is not 0.385 J/gK. (If it was polystyrene, which absorbs/ transmits little heat, the effective value of C would be very low, reducing L.)

Using +/- 40 is best.

However, the spread in the actual results is much smaller

* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.

<h3>Other sources of error: </h3>

L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?

* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt -which explain small values of n

* some water might have been left in container when unmelted ice was weighed (eg clinging to ice) - again this could explain small n;

* poor insulation - heat gained from surroundings, melting more ice, increasing n - but this would reduce measured L below 334 J/g not increase it.

* calorimeter still cold from last trial when next one started, not given time to reach same temperature as water - this would reduce n.

3 0
3 years ago
The fractured surface of which moon suggests a worldwide ocean of liquid water beneath its icy surface
dimaraw [331]
I think your answer is pluto...
5 0
3 years ago
Which of the following is a dihybrid cross?a. RrMM Rrmmb. RRMM rrmmc. RrMm RrMmd. rrMM RRmme. RrMm rrmm
Minchanka [31]

Answer:

c. RrMm x  RrMm

Explanation:

A dihybrid genotype is the one that is heterozygous for two genes. Hence, a dihybrid cross is a cross between two individuals that are hybrid for alleles of two different loci.

Among the given examples, the cross RrMm x RrMm is a dihybrid cross. Here, both given genotypes are heterozygous or hybrid for two loci (both genotypes have one dominant and one recessive allele for both the genes under study, R and r; M and m).

This cross explains the inheritance of two loci or two genetic traits. Hence, it is an example of dihybrid cross.

3 0
2 years ago
The fluid mosaic model describes the plasma membrane as consisting of
Hoochie [10]
The answer is e, as in terms of d, proteins are not sandwiched between the two bilayers, but wedged in the bilayer to selectively let in different molecules. With c, phospholipids do not drift, into the membrane, for it is the membrane itself. With b it is the other way around, the phospholipid bilayer makes up the membrane, and proteins are embedded into it. Lastly, for a, a fluid mosaic model wouldn’t portray phospholipids drifting in the dark phospholipid bilayer, as they are an independent molecule consisting only as the cell membrane. Hope that’s helpful! :)
6 0
2 years ago
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