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dolphi86 [110]
3 years ago
14

During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima

tely 103 cm/s. The blood volume traverses the aortic arch, exiting with the same speed but opposite direction. Assume the mass density of the blood is 1060 kg/m3 blood, the aortic arch remains stationary, and that the heart rate is 59 bpm. What is the average force exerted by the blood on the wall of the aorta.
Physics
1 answer:
vova2212 [387]3 years ago
8 0

Answer:

The value  F  =  0.1396 \ N

Explanation:

From the question we are told that

   The volume blood  ejected is  V  =  65 \ cm^3  = 65*10^{-6} \  m^3

    The velocity of the blood ejected is  v  = 103 \  cm/s  = \frac{103}{100} = 1.03 \ m/s

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

     The heart beat is R = 59 \  bpm(beats \  per \  minute) = \frac{59}{60}= 0.9833\ bps

The average force exerted by the blood on the wall of the aorta is mathematically represented as

      F  =  2 * \rho  *  V  *  R *  v

=>    F  =  2 * 1060  *  65*10^{-6}  *  0.9833 *  1.03

=>    F  =  0.1396 \ N

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almond37 [142]

AS

work done =W = F.d = F d cosФ     (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body.  F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example:  A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done  dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0   ( as cos 90°=0)

5 0
3 years ago
Death Star has a diameter of 160,000m and a mass of 5.1e17kg. Millennium Falcon has a mass of 1.36e6kg (data from Wookieepedia)
lions [1.4K]

Answer:

7229 N

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is given by:

F=G\frac{mM}{R^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M=5.1\cdot 10^{17} kg is the mass of the Death Star

m=1.36\cdot 10^6 kg is the mass of the Millennium Falcon

R=\frac{160,000 m}{2}=80,000 m is the radius of the Death Star

Substituting numbers into the equation, we find the force

F=(6.67\cdot 10^{-11})\frac{(1.36\cdot 10^6 kg)(5.1\cdot 10^{17} kg)}{(80,000 m)^2}=7229 N

5 0
3 years ago
Using the results of Question 1 that would apply if the collision were inelastic, compute (using Excel in the yellow highlighted
ch4aika [34]

Answer:

The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.

The lost energy became heat and sound energy.

Explanation:

During inelastic collision, the kinetic energy of a moving object does not conserve. It changes into another form of energy such as sound energy and heat energy etc.

For example, when a moving car hit another car or wall etc, the kinetic energy is converted into sound and heat energy. This type of collision is inelastic collision.

4 0
3 years ago
3. What is the difference between Synthetic Oil, Semi-Syntheic oil (Blend), High milage motor oil, and conventional motor oil? ​
MariettaO [177]

Answer:

When evaluating synthetic blends, it's helpful to define the terms “synthetic blend” and “semi-synthetic”. Generally speaking, synthetic blends and semi-synthetic refer to the same thing: an oil that uses a combination of conventional and synthetic base oils in its formulation.

Explanation:

6 0
3 years ago
PE=30J, m=?, g=10m/s2, h=10m
OleMash [197]
Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

PEgrav=mgh
30J=m(10)(10[tex] \frac{30}{100} =m)[/tex]

Do operations that you can with what is given first.

30J=m(100m)

Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

\frac{30}{100} =m

m = 0.30kg

5 0
3 years ago
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