The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.How many capillaries are there?

Answers

Answer 1

Answer:

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

$A_1v_1=nA_2v_2$ (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

$A=\pi r^2\n\nA_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\n\nA_2=\pi r_2^2=\pi (5*10^(-4)cm)^2=7.85*10^(-7)cm^2$

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

$n=(A_1v_1)/(A_2v_2)=((3.1415cm^2)(40cm/s))/((7.85*10^(-7)cm^2)(0.10cm/s))=1.6*10^9$

Then, the number of capillaries is 1.6*10^9

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Convert 56km/h to m/s.

Answers

Explanation:

15.556 metres per second

Help!!! Need answer ASAP.

Answers

Answer:

Hey

Explanation:

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }$

t is observed time.

$t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

The energy provided each hour by heat to the turbine in an electric power plant is 9.0×10^12 J. If 5.4 × 10^12 J of energy is exhausted each hour from the engine as heat, what is the efficiency of this heat engine?

Answers

Answer:

60 %

Explanation:

Efficiency is defined as the ratio of output power to the input power.

Input energy each hour = 9 x 10^12 J

Output energy each hour = 5.4 x 10^12 J

Efficiency = Output energy per hour / input energy per hour

Efficiency = (5.4 x 10^12) / ( 9 x 10^12) = 5.4 / 9 = 0.6

Efficiency in percentage form = 0.6 x 100 = 60 %

Final answer:

The efficiency of a heat engine is calculated using the formula (Energy Input - Energy Output) / Energy Input. Given the figures, the efficiency of the engine is 40%, indicating that 40% of the input energy is converted into work.

Explanation:

The efficiency of a heat engine is determined by the ratio of work output to energy input. In the given scenario, the turbine in an electric power plant is supplied with 9.0 x 10^12 joules of energy, and 5.4 x 10^12 joules of energy is expelled as heat per hour. We can calculate the efficiency using the equation:

Efficiency = (Energy Input - Energy Output) / Energy Input

By substituting the given values, Efficiency = (9.0 x 10^12 J - 5.4 x 10^12 J) / 9.0 x 10^12 J = 0.4 or 40%

This means the heat engine of the power plant has a 40% efficiency, meaning 40% of the energy input is converted into work while 60% is discarded as waste heat.

Learn more about Thermal Efficiency here:

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A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β