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rosijanka [135]
3 years ago
8

Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing thro

ugh the aorta (in grams of blood per second)?
Physics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

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Technician A says that the freezing point of a substance is the temperature at which the substance changes state from a solid to
pashok25 [27]

Answer:

Technician B is correct

Explanation:

Freezing is a method of conversion of substance in its liquid state to solid state. It is the process by which a liquid substance changes to a solid at a particular temperature.

Increasing the pressure and decreasing the temperature of a liquid increases its freezing point. For example, in other to freeze water i.e to change water to ice, it has to be kept in a fridge at a temperature lower than the temperature of the water. The essence of covering the fridge after placing the water in the fridge is to increase the pressure of the liquid hence increasing its freezing rate.

Based on the above explanation, it can be concluded that technician B is correct.

5 0
3 years ago
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0
2 years ago
An ice chest at a beach party contains 12 cans of soda at 4.05 °C. Each can of soda has a mass of 0.35 kg and a specific heat ca
BARSIC [14]
Below is the solution:

Heat soda=heat melon 
<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
<span>43260t=873600 </span>
<span>t=873600/43260 </span>
<span>t=20.19 deg celcius</span>
4 0
3 years ago
You throw a ball upward with an initial speed of 4.3 m/s. When it returns to your hand 0.88 s later, it has the same speed in th
djverab [1.8K]

Answer:

The acceleration is -9.8 m/s²

Explanation:

Hi there!!

When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.

The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:

acceleration = final speed - initial speed/ elapsed time

acceleration = -4.3 m/s - 4.3 m/s / 0.88 s

acceleration = -9.8 m/s²  

4 0
3 years ago
The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0o to
Allushta [10]

Answer:

0.572 Hz

Explanation:

given,

length of simple pendulum, l = 0.76 m

mass of the bob, m = 365 g = 0.365 Kg

angle made from the vertical, = 12°

frequency, f = ?

f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}

f = \dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{0.76}}

f = \dfrac{1}{2\pi}\times 3.59

       f = 0.572 Hz

The frequency at which pendulum vibrates is equal to 0.572 Hz

3 0
3 years ago
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