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2 years ago

All of the following are factors affecting flow rate except what?

Physics

1 answer:

faust18 [17]2 years ago

5 0

Answer:

c. Vessel side holes

Explanation:

The "Poiseuille formula" which is given by $\\\begin{aligned} \small Q& = \small \frac{\pi r^4}{8 \eta}.\frac{\Delta P}{\Delta L}\\\end{aligned}$ describes the volumetric flow rate ( $\small Q$ ) through tubular sections.
Here, $\Delta P,\,\, \Delta L,\,\, r,\,\, \eta$ represent the injection pressure difference, the length of the section, the radius of the section and the viscosity index of the fluid that flows through the section respectively.
With this, one can confirm that all the factors except the vessel side holes affect the flow rate.
Side holes, however, are a factor that could give a measure of how much volume would flow to a particular location. In such a situation the flow rate remains unchanged and one location would receive a lower volume (not the whole) as some volume would spill out at the side holes.

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Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls

Dmitrij [34]

Answer:

$(a) 189.23 N$ , $(b) 47.31 N$ and $(c) 141.92 N$ .

Explanation:

Three balls are shown in figure having charge $q=1.45 \mu C$ . The middle ball, $B$ , is positively charged having charge $+q$ , and the remaining two outside balls, $A$ and $C$ , are negatively charged having charged $-q$ as shown.

$AC=20 cm$ and $AB=BC=10$ cm as B is the mid-point of AC.

Let $d_1=AC=20\times 10^{-3}m$ and $d_2=AB=BC=10\times 10^{-3}m$

From Coulomb's law, the magnitude of the force, $F$ , between two point charges having magnitudes $q_1 \& q_2$ , separated by distance, $d$ , is

$F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)$

where, $\epsilon_0$ is the permittivity of free space and

$\frac {1}{4\pi\epsilon_0}=9\times 10^9$ in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

$F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}$

$\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}$

$\Rightarrow F_{AB}=189.23 N$

(a) The magnitude of the repulsive force between balls A and C is

$F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}$

$\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}$

$\Rightarrow F_{AC}=47.31 N$

$F_{net}=189.23-47.31 N$

$\Rightarrow F_{net}=141.92 N$

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3 years ago

A movie stunt performer is filming a scene where he swings across a river on a vine. The safety crew must use a vine with enough

Julli [10]

Answer:

1125.66956 N

Explanation:

m = Mass of stunt performer

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of the swing = 7 m/s

T = Tension

r = Radius of the swing = Length of vine = 11.5 m

From the free body diagram

$T-mg-m\frac{v^2}{r}=0\\\Rightarrow T=mg+m\frac{v^2}{r}\\\Rightarrow T=m(g+\frac{v^2}{r})\\\Rightarrow T=80(9.81+\frac{7^2}{11.5})\\\Rightarrow T=1125.66956\ N$

The minimum tension force the vine must be able to support without breaking is 1125.66956 N

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3 years ago

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

Likurg_2 [28]

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

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3 years ago

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Describe the relationship between joules, meters, and newtons

Naily [24]

Newton is a unit of force. Joules is an amount of work which is equal to the force times distance, or newton meters. So the product of newtons and meters makes joules.

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3 years ago

Someone answer? Please

dexar [7]

I don’t know ask your parents

6 0

2 years ago