1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ilya [14]
3 years ago
11

Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various

factors, calculate the new flowrate for the following changes with all other factors remaining the same as in the original conditions: A new fluid with 6.00 times greater viscosity is substituted. Poiseuille flow is given by:
Physics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

Q_{2}=1200cm^{3}/s

Explanation:

Given data

Q₁=200cm³/s

We know that:

F=n\frac{vA}{l}\\

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s

You might be interested in
I WILL MARK YOU THE BRAINLIEST
mr Goodwill [35]

Explanation:

700N right

to get the net force

you gotta let one direction be the negative ( the smaller force)

so the total force towards the left is 100N ( 60 + 40= 100)

which is smaller than the right force which is 800 N so you let 100 N be negative

so without even calculating , you can know that it will be moving towards the right because right force > left force

your add both forces ( remember 100 N is negative)

so 800N + ( - 100N)

= 700N

towards the right

hope this helps

this is just one method that helped me understand

please mark it brainliest

3 0
3 years ago
Read 2 more answers
) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe
pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0
3 years ago
Read 2 more answers
A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241
Ivahew [28]

Answer:

ω = 3.61 rad/sec

Explanation:

Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.

μmg = mv^2/r = mω^2r

Thus;

μg = ω^2r

ω^2 = μg/r

ω = √(μg/r)

ω = √(0.321 * 9.8)/0.241

ω = √(13.05)

= 3.61 rad/sec

3 0
3 years ago
Why do we see the sun before it actually rises in the morning?
Lera25 [3.4K]

Answer:

It appears to rise and set because of the Earth's rotation on its axis. It makes one complete turn every 24 hours. It turns toward the east. As the Earth rotates toward the east, it looks like the sun is moving west.

Explanation:

6 0
3 years ago
How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?​
miss Akunina [59]

Answer:

The answer to your question is    Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

           Pe = mgh

Substitution

           Pe = (50)(5)(9.81)

Simplification

           Pe = 2452.5 J

8 0
3 years ago
Other questions:
  • A force does 30000 J of work along a distance of 9.5m. Find the applied force.
    8·1 answer
  • A brass bar, density 9.87g/cm3, has a volume of 20.25cm3. What is the mass of this brass bar?
    8·1 answer
  • Two identical stars with mass M orbit around their center of mass. Each orbit is circular and has radius
    10·1 answer
  • Raising the temperature of a gas will increase its pressure if which of the following happens?
    6·2 answers
  • The above Free Body Diagram represents the motion of a toy car across a floor from left to right. The weight of the .5 kg car is
    12·1 answer
  • When different resistors are connected in parallel across an ideal battery, we can be certain that: a) their equivalent resistan
    11·1 answer
  • The speed of light changes as it passes from one medium to the next. In which image does the speed of light change the most?
    8·2 answers
  • The police department is excited to have some new motorcycle units. One officer said that these motorcycles can go from 0 miles
    10·1 answer
  • A 500 kg satellite experiences a gravitational force of 3000 N, while moving in a circular orbit around the earth. Determine the
    9·1 answer
  • How many electrons flow through a 3 A current in 3 seconds
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!