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3 years ago

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Jeff's body contains about 5.12 L5.12 L of blood that has a density of 1060 kg/m3.1060 kg/m3. Approximately 45.0%45.0% (by mass)

of the blood is cells and the rest is plasma. The density of blood cells is approximately 1125 kg/m3,1125 kg/m3, and about 1%1% of the cells are white blood cells, the rest being red blood cells. The red blood cells are about 7.50????m7.50μm across (modeled as spheres). What is the mass of the blood mbloodmblood in Jeff's body?

1 answer:

hjlf3 years ago

4 0

Answer:

5.4272 kilogram is the mass of the blood in Jeff's body.

Explanation:

Mass of the blood in Jeff's body = m

Volume of the blood in Jeff's body = V = 5.12 L

Density of blood = $1060 kg/m^3=1.060 kg/L$

$1m^3= 1000 L$

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

$D=\frac{M}{V}$

$M=D\times V$

$M=1.060 kg/L\times 5.12 L=5.4272 kg$

5.4272 kilogram is the mass of the blood in Jeff's body.

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X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

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3 years ago

Which graph below represents how the velocity of the sphere changes over time when falling with constant acceleration?

Amiraneli [1.4K]

Graph B represents the velocity of the sphere changes over time when falling with constant acceleration.

Acceleration is the measure of how quickly a body's velocity varies with regard to time, and constant acceleration occurs when a body's velocity changes proportionately over a period of time, or at a constant rate. It measures in m/s2.

It is claimed that a body has continual positive acceleration when it begins to move with an initial velocity of zero and gradually increases to a positive value over time.

Constant positive acceleration is demonstrated by a ball falling freely in a vertical direction.

To know more about constant acceleration. visit : brainly.com/question/9754169

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1 year ago

A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend

liq [111]

The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

   (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.

We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From y=0 to y=2.40 is a distance of 2.40 upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

Work = integral of (F·dy) evaluated from 0 to 2.40

= integral of (-2.85 y² dy) evaluated from 0 to 2.40

   = (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .

Now, integral of (y² dy) = 1/3 y³ .

Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)

      = 1/3 · 13.824 = 4.608 .

And the work = (-2.85) · the integral

   = (-2.85) · (4.608)

   =    - 13.133 .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.

-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.

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3 years ago

2. What is the standard value of acceleration due to gravity or "g"?

svp [43]

Answer:

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2 (about 32.17405 ft/s2).

Explanation:

6 0

3 years ago

Read 2 more answers

What would the speed of this wave be?

valentinak56 [21]

Answer:

c) 100,000 m/s

Explanation:

You need to take the same wave length from the top graph and bottom one, so let's take half a wave length then in the top one that is 0.005, but in the bottom one it's 2000/4 = 500 because they are smaller and there are 4 half waves before you get to 2000, whereas in the top one there is 1 half wave before you get to 0.005 on the graph.

Now use speed = distance / time

speed = 500 / 0.005 = 100 000 m/s

7 0

3 years ago