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Kipish [7]
3 years ago
9

Vascular resistance is the friction force that opposes blood flow in a blood vessel. what two factors are the most important in

generating friction between blood and the walls of a vessel?
Physics
1 answer:
makvit [3.9K]3 years ago
6 0
The two most important factors are Systematic Vascular Resistance as and Vascular Resistance.
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Which is occurring when work is being done?
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Answer:

When work is being done, the practie of labor is being performed. When work isn't being completed, this is a sign of procrastination.

Explanation:

In the answer.

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The amount of energy required to raise the temperature of 1 kilogram of a substance by 1 Kelvin is called its
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Specific heat.  The definition of specific heat is the amount of energy required to raise the temperature of 1g of a substance by 1K or 1°C.
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A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r
Murrr4er [49]

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l =  4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s

Let \omega is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s

So, the maximum angular speed of the ball is 2.1 rad/s.

4 0
3 years ago
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0
3 years ago
Which feature of a longitudinal wave corresponds to a trough in a transverse wave?
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It’s the crest, the crest is the top part of the wave and the trough is the bottom so they correspond
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