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1 month ago

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A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,

what is the total distance traveled by the red spot?
a.) 4A
b.) 1/2 A
c.) 1/4 A
d.) A
e.) 2A

1 answer:

serg [3.5K]1 month ago

6 0

Answer: E. 2A

Explanation:

The amplitude (A) of a wave refers to the maximum displacement from its rest position. The peak value of the amplitude corresponds to the crest, while the lowest value represents the trough.

For a wave to complete a full cycle, it must travel back and forth from its displacement point.

<pIf the red-marked spot on the string completes one cycle, it indicates that it has traversed through the crest and the trough, which amounts to 2 amplitudes.

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Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

ValentinkaMS [3465]

Answer:

a) Blood mass is $m= 5.7876kg$

b) The count of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the problem statement, we learn that

The blood volume is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of blood is $\rho_b = 1060 kg/m^3$

% of blood which consists of cells is = 45.0%

the % of blood that is plasma is = 55.0%

density of blood cells is $\rho_d = 1125kg/m^3$

% of cells that are white is = 1%

% of cells that are red is = 99%

The red blood cell diameter is = $7.5 \mu m = 7.5*10^{-6}m$

The red blood cell radius is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

The mass is generally represented mathematically as

$m = \rho_b * V_b$

Substituting values

$m = 1060 * 0.00546$

$m= 5.7876kg$

Cell mass is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The white blood cells volume is $V_w$ = 1% of the cells volume

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The red blood cells volume is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The total red blood cell count is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The total white blood cell count is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The overall number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

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A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat

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The focal length of the lens while in water is noted to be 150 cm, whereas in air, it measures 60 cm. To derive these values, the formula incorporates the variations in the refractive index of glass compared to that of the surrounding medium.

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Which of these nebulae is the odd one out?

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The correct choice is D!

Clarification:

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Determine the mass of a ball with a velocity of 40.0 m/s and a wavelength of 8.92 Ã 10-34 m.

Sav [3153]

The wavelength can be calculated as Planck's constant divided by the momentum of the ball.
This translates to:
lambda = h / p.............> equation I
Momentum is equal to mass times velocity............> equation II

By substituting equation II into equation I, we obtain:
lambda = h / mv
Here are the values provided:
lambda = 8.92 * 10^-34 m
Planck's constant = 6.625 * 10^-34
velocity = 40 m/sec

Substituting these values into the previous equation, we calculate the mass as follows:
8.92*10^-34 = (6.625*10^-34) / (40*m)
mass = 0.0185678 kg

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14 days ago

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the

kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

$F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N$

In the case of the N-H-N combination:

The total force acting from this combination is:

$F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N$

The force that thymine applies on adenine is:

$F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N$

When rounded to three significant figures, the net force is $8.86\times 10^{-9}N$

b)

The negative value indicates that the force is attractive, as it is aimed towards thymi.

c)

The force acting on the electron due to the proton is:

$F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N$

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

d)

The ratio of the above forces is:

$\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3$

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

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1 month ago