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2 years ago

8) What kind of circulation carries blood to all organs and systems?

Physics

2 answers:

vivado [14]2 years ago

8 0

The heart pumps the blood

Roman55 [17]2 years ago

5 0

There isn’t one blood circulatory system in the human, but two, which are in fact connected. The systemic circulation provides organs, tissues and cells with blood so they get oxygen and other vital substances. The pulmonary circulation is where the fresh oxygen we breath in enters the blood.

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A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

2 years ago

81. A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a

Irina18 [472]

Answer:

Explanation:

Given

mass of squirrel $m=560 gm$

Surface area of squirrel $A=930 cm^2$

and the area which face $A_f=\frac{A}{2}=465 cm^2$

height of tree $h=5 m$

Coefficient of drag $C=1$

drag Force $F_d=\frac{1}{2}C\cdot \rho \cdot A\cdot v^2$

Terminal velocity is given

$F_d=mg$

$\frac{1}{2}C\cdot \rho \cdot A\cdot v^2=mg$

$v=\sqrt{\frac{2\times m\times g}{\rho \times C\times A_f}}$

$v=\sqrt{\frac{2\times 0.560\times 9.8}{1.2\times 1\times 465\times 10^{-4}}}$

$v=13.9 m/s$

(b)Mass of person $m=56 kg$

$v^2-u^2=2gh$

$u=0$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 5}$

$v=9.89 m/s$

7 0

3 years ago

Read 2 more answers

List examples of two energy transformations- one produced by an electric and the other by a mobile phone.

Phantasy [73]

Answer:

an electric or landline turns electricity into mechanical energy while a mobile phone turns electricity to chemical to heat

Explanation:

3 0

3 years ago

a spotted lizard runs at 3m/s at top speed. a girl wants to catch the lizard to keep as a pet. where should the girl place her c

Ann [662]

The acceleration due to earth's gravity is -9.8 m/s [dn] I thought... I'm assuming this is a projectile motion question asking for the range.

Break each kinematic quantity into their x and y components.

x y

v₁ = 3 m/s 0

v₂ = 3 m/s ?

Δd = ? -1.5

Δt = ? ?

a = 0 -10 m/s²

So the variable we are trying to find is Δdx (x component of displacement). We need to use a kinematic equation to do so. However, we obviously don't have enough given to find Δdx. This means we need to find something first, something we can use. How about Δt? Δt can be applied to both the x and y components. We have enough information in the y component list to find Δt. We can use this formula and solve for Δt.

Δdy = v₁y ( Δt ) + 1/2 ( ay ) ( Δt )²

Δdy = 1/2 ( ay ) ( Δt )² <- the first term cancels out since v₁y = 0.

2Δdy = ay ( Δt )²

2Δdy / ay = ( Δt )²

√ 2Δdy / ay = Δt

√ 2(-1.5 m/s) / -9.8 m/s² = Δt

√ -3.0 m/s / -9.8 m/s² = Δt

√ 0.306122449 s² = Δt

0.5532833352 s = Δt

Now, we can use this newly found quantity to solve for Δdx using the x component values using the appropriate kinematic equation.

Δdx = ( v₁x + v₂x / 2) ( Δt )

Δdx = ( ( 3.0 m/s + 3.0 m/s ) / 2 ) ( 0.5532833352 s )

Δdx = ( 6.0 m/s / 2 ) ( 0.5532833352 s )

Δdx = ( 3.0 m / s )( 0.5532833352 s )

Δdx = 1.659850006 m

Therefore, the girl should place her cage 1.7 m away from the platform to catch the lizard.

This solution assumes that the acceleration due to gravity is -10 m/s² [dn] and not -9.8 m/s² [dn]. If you need -9.8 m/s² [dn], then just substitute it into my solution. This was a pain to type lol

6 0

3 years ago

How does technology or research of a nuclear physicist involve electromagnetic or sound waves

kap26 [50]

Answer: While quantum physics is usually concerned with the basic building blocks of light and matter, for some time scientists have now been trying to investigate the quantum properties of larger objects, thereby probing the boundary between the quantum world and everyday life. For this purpose, particles are slowed down with the help of electromagnetic waves and the motional energy is drastically reduced. Therefore, one also speaks of "motional cooling."

Quantum properties occur when particles are cooled to their fundamental quantum ground state, that is to the lowest possible energy level. While so far the only way to cool particles to the ground state has been to make them interact with photons trapped in an electromagnetic resonator, theoretical physicists led by Carlos Gonzalez-Ballestero and Oriol Romero-Isart from the Department of Theoretical Physics at the University of Innsbruck and the Institute of Quantum Optics and Quantum Information (IQOQI) of the Austrian Academy of Sciences in collaboration with experimentalist Jan Gieseler from Harvard University and ICFO in Barcelona now propose to make the motion of magnetic particles interact with the internal acoustic waves that are confined inside every particle.

Explanation:

Sound waves in micro-magnets

In analogy to photons -- the quanta of light -- vibrations in a solid body can be described as so-called phonons. These small sound wave packets propagate through the crystal lattice of the solid. "The phonons are very isolated and interact with the movement of the particle motion only through magnetic waves," explains Carlos Gonzalez-Ballestero. "In our work we now show that this interaction can be controlled by a magnetic field." This allows to realize quantum experiments without photons, and therefore even with light-absorbing particles. "Conversely, we also show that the strong interaction between motion and phonons provides a path to probe and manipulate the elusive and exotic dynamics of acoustic and magnetic waves in very small particles," adds Oriol Romero-Isart. The new method also opens up new possibilities for quantum information processing, for example, by using phonons as a quantum memory.

4 0

3 years ago