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Drupady [299]
3 years ago
5

The influence of blood vessel diameter on peripheral resistance is ________.

Physics
1 answer:
NeX [460]3 years ago
6 0

Answer:

The influence of diameter of the blood vessel on peripheral resistance is significant because resistance is inversely proportional to the fourth power of the diameter.

Explanation:

The influence of diameter of the blood vessel on peripheral resistance is significant because the relation between the peripheral resistance and the diameter is given as, resistance is inversely proportional to the fourth power of the diameter. Thus, with small increase or decrease in the value of diameter, the peripheral resistance may vary by a significant amount.

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what is the density of a material if its mass is 450 kilograms and occupies a volume of 25 centimeters cubed​
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A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
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(a) 4.0 s

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a=\frac{v-u}{t}

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u is the initial velocity

t is the time interval

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u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

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The deceleration is given by the problem, and its value is -5.0 m/s^2.

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F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

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