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Liono4ka [1.6K]
3 years ago
10

If an ideal fluid element moves from a region of lower pressure to a region of higher pressure, is the work done on it by the pr

essure force positive, negative, or zero?
Physics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

Positive

Explanation:

An ideal fluid is on which is incompressible and non-viscous.

In an ideal fluid element when a fluid element moves from a region of high pressure to region of low pressure then the work done on the fluid element is positive because the force acts in the direction from high pressure towards low pressure and displaces the fluid element, so the work-done is positive.

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A car is traveling at 7.1m/s accelerates 4.3 m/s^2 to reach a speed of 16.0 m/s how long does it take for this acceleration to o
Bond [772]

<u>Answer</u>

2.06977 seconds


<u>Explanation</u>

Acceleration, a, is the rate of change of velocity

a = (v-u)/t

  where v and u are final and initial velocities respectively.

4.3 = (16.0 - 7.1)/t

4.3 = 8.9/t

t = 8.9/4.3

   = 2.06977 seconds

4 0
3 years ago
Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to
Nataliya [291]

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1}  =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}

As we know, P1 must be equal to \frac{F_{1} }{A_{1}}, so, replacing P1 in the equation, we have:

P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})

And

F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as Q = V A, where Q is flow, V is velocity and A is area. As the same, we can write this: Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}. In the last two equations, let's clear Velocities.

V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

3 0
3 years ago
The splitting of an atom is called Blank Space __________.
horsena [70]

Fission is the answer

8 0
3 years ago
if change in blood pressure between the brain and the feet is 1.88×10^4 pa .what will be the volume flow rate from head to feet
klio [65]

Answer: 3765.66 \frac{m^{3}}{s}

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

Q=\frac{\pi r^{4}\Delta P}{8\eta L}

Where:

Q  is the Volume flow rate

r=23 cm \frac{1 m}{100 cm}=0.23 m  is the effective radius

L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m  is the length

\Delta P=1.88(10)^{4} Pa  is the difference in pressure

\eta=3(10)^{-3} Pa.s is the viscosity of blood

Solving:

Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}

Q=3765.66 \frac{m^{3}}{s}

4 0
4 years ago
A uniform plank is 3.0 m long and has a mass of 10 kg. It is secured at its left end in a horizontal position to be used as a di
garik1379 [7]

To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.

By torque it is understood that

\tau = F*d

Where,

F= Force

d = Distance

The value of the given Torque acts from the center of mass causing it to rotate clockwise.

The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.

I enclose a graph that allows us to understand the problem in a more didactic way.

The correct answer is D.

7 0
3 years ago
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