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3 years ago

Figure 15-1 is an example of heat transfer by ____________________.

Biology

2 answers:

Setler79 [48]3 years ago

8 0

The right answer is Convection.

Convection is one of three modes of heat transfer with conduction and radiation. The term convection refers to the heat transfer occurring between a surface and a moving fluid when these are at different temperatures. In addition to the energy transfer due to diffusion, there is also transfer through the movement of the fluid. The latter is associated with the fact that multiple molecules have a collective motion, which implies a heat transfer in the case where there is a thermal gradient.

alexira [117]3 years ago

6 0

Convection is the answer

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Answer:

The correct answer is a cancerous tumor.

Explanation:

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3 years ago

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Answer:

D. genetic material ..........

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The state of health and functioning of the liver is often assessed with dye-tracer techniques. The dye used most frequently is b

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Answer

1.Radioactive or chemical decay

2. X(t) = Xoe^-kt

lnY(t) = -t + lnYo

4. 6.07mg

Explanation:

Let the liver and and blood compartment be represented by the symbol L and B respectively

For the liver

Suppose a first Order removal process started with an amount X, in which amount b disappeared in time t, the process is decay process which can be represented as follows,

∫dX/dt = -K1X

By rearrangement and integration;

∫dX/X = -K1t

ln X = -K1t + C

Since At t= 0, X = Xo then

C = lnXo

The equation becomes:

lnX = -K1t + lnXo

lnX - lnXo = -K1t

ln(X/Xo) = -K1t

X/Xo = e^-kt

X(t) = Xoe^-kt

X(t) = Xoe^-0.5t........(2)

Similarly for B (blood), suppose a first order flow flow of the dye move from the blood to the liver, let Y be the initial concentration, and amount b that has flown to the liver in time t

B---------> B(t)

t=0 Yo 0

t=t. Y-b. b

dY/dt = -K12(Y-b)...........(3)

Let Y-b = Y(t)

∫dY/dt = -∫K12t

By rearrangement and integration;

∫dy/Y(t) = ∫-K12dt

lnY(t) = -K12t + C1

at t= 0, C1 = ln(Yo)

Therefore ln Y(t) = -K12t + lnYo

But K12 =1

ln Y(t) = -t + lnYo.............(4)

(3) The assumptions used here

is that of a decay for the liver . The amount remaining taking as the amount of a substance

(4) using the equation 2,

X(t)= Xo e^-K1t........(2)

For time t = 1hour, and an initial amount X = 10mg, K = 0.5

X(t) = 10× e^-0.5

X(t) = 10 × 0.607

X(t) = 6.07mg

(5) within the scope of information presented, I have no data to make this judgment.

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3 years ago