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scoray [572]
3 years ago
8

In a test on a 6 "diameter light oxidation cast iron conduit where the flow rate was 26.5 L / s, the pressure was measured at tw

o points A and B, distant from 1017 m where values of 68.6.104N / m2 and 20.6.104N / m2 respectively. Considering that the difference in quota between A and B was 30m and the quota of A being lower than B.
Determine: the speed of water in the pipe?
Physics
1 answer:
saul85 [17]3 years ago
6 0

Look kid, this question has been up for 16 hours think its time to let it go......... hoped i helped. ✔verified✔

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A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦
bija089 [108]
At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
pV=nRT
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and R=0.082 atm L K^{-1} mol^{-1} the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
n= \frac{pV}{RT}= \frac{(1.00atm)(20.0L)}{(0.082 LatmK^{-1}mol^{-1})(273 K)}=0.89 mol
5 0
3 years ago
What happens when two forces act in the same direction?
Wewaii [24]

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

5 0
3 years ago
Read 2 more answers
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
jok3333 [9.3K]

Answer:

S_a_v_e_r_a_g_e=48km/h

Explanation:

Ok, the average speed can be calculate with the next equation:

S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}   (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

Total\hspace{3}distance=2*d

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

Total\hspace{3}time=t1+t2

From basic physics we know:

t=\frac{d}{S1}

so:

t1=\frac{d}{S1}

t2=\frac{d}{S2}

Using the previous information in equation (1)

S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }

Factoring:

S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}   (2)

Finally, replacing the data in (2)

S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h

5 0
3 years ago
A ___________ will help you plan the rest of your life around this fitness priority. A. coach B. fitness goal C. fitness evaluat
tatiyna

Answer:

B. fitness goal I believe. please like this, I need points. ty!

Explanation:

8 0
2 years ago
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