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3 years ago

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A 7.1 cm diameter horizontal pipe gradually narrows to 5.4 cm . When water flows through this pipe at a certain rate, the gauge

pressure in these two sections is 32.5 kPa and 20.6 kPa , respectively. What is the volume rate of flow?

1 answer:

crimeas [40]3 years ago

4 0

Answer:

Q=22.6L/s

Explanation:

First you must consider the continuity equation at points 1 and 2, which indicates that both flows are of equal value, in this way you get an equation between the two flow rates.

Then you raise the Bernoulli equation taking into account that the height is the same, which makes the term h1-h2 zero.

Using the equations above to calculate one of the speeds.

Finally you find the flow by multiplying the speed by the area.

I attached procedure

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Help please! this is physics !

KATRIN_1 [288]

Answer:

4. The choose b. 0.000355

Ans; 3.55× 10-⁴ = 0.000355

5. The choose C. 80600

Ans; 8.06 ×10⁴= 806×10² = 80600

I hope I helped you^_^

4 0

3 years ago

Read 2 more answers

The temperature inside my refrigerator is about 4 degrees C. If I place a balloon in my fridge that initially has a temperature

maksim [4K]

v

Convert the given temperatures from celsius to kelvin since we are dealing with gas.

To convert to kelvin, add 273.15 to the temperature in celsius.

T1 = 22 + 273.15 = 295.15 k

T2 = 4 + 273.15 = 277.15 k

V1 = 0.5 L

Let's find the final volume (V2).

To solve for V2 apply Charles Law formula below:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

5 0

1 year ago

Match Newton's laws with their descriptions. 1. Every applied force is opposed by an equal force. 2 .A force must be applied to

Marina CMI [18]

A would be number 2. Newton's First Law states that an object at rest, will stay at rest and an object in motion, will stay in motion, unless acted upon by an unbalanced force. B would be number 3. His Second Law states that <span>the sum of the forces acting on a body is equal to the product of the mass of the body and the acceleration produced by the forces. And, C would be number 1. His Third Law states that for every action, there is an equal and opposite reaction. Hope this helps!</span>

6 0

4 years ago

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

vova2212 [387]

Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s, 8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago