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muminat
3 years ago
9

a piping system consists of 100 ft of 2-inch pipe, a sudden expansion to 3-inch pipe, and then 50 ft of 3-inch pipe. Water is fl

owing at 100 gal/min through the system. What is the pressure difference from one end of the pipe to the other
Physics
1 answer:
ikadub [295]3 years ago
8 0

Answer:

16+15+19= ??

Am just messign with u lol

Explanation:

anwser s 19 inches

i

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What is a pickel in a nasty way do?
katen-ka-za [31]

Answer:

pickle rick

Explanation:

8 0
3 years ago
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until a train is a safe distance from the station, it must travel at 5 m/s. once the train is on open track, it can speed up to
MatroZZZ [7]

Answer:

5 meters per second squared

Explanation:

We calculate the acceleration using the formula:

a = (vf - vi) / t

where "vf" is the final velocity, "vi" the initial velocity, and "t" the time it took to change from the initial velocity to the final one.

In our case:

a = (45 - 5) / 8 = 40 / 8 = 5 m/s^2

4 0
4 years ago
A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respecti
Morgarella [4.7K]

Answer:

The value of the effective (rms) voltage of the applied source in the circuit is 132 V

Explanation:

Given;

effective (rms) voltage of the resistor, V_R = 65 V

effective (rms) voltage of the inductor, V_L = 140 V

effective (rms) voltage of the capacitor, V_C = 80 V

Determine the value of the effective (rms) voltage of the applied source in the circuit;

V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V

Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.

6 0
4 years ago
How do you find average velocity (average) from acceleration) and time (t)?
Tasya [4]

Average velocity is defined as the ratio in change in position to change in time,

v[ave] = ∆x/∆t

which on its own doesn't have anything to do with acceleration.

<u>If acceleration is constant</u>, the average velocity is the literal average of the initial and final velocities,

v[ave] = (v[final] + v[initial]) / 2

If this constant acceleration has magnitude a, the final velocity can be expressed in terms of the initial velocity by

v[final] = v[initial] + a*t

and plugging this into the previous equation gives

v[ave] = (v[initial] + a*t + v[initial])/2

v[ave] = v[initial] + 1/2*a*t

If the body in consideration is <u>initially at rest</u>, then

v[ave] = 1/2*a*t

which might be the relation you're looking for. But bear in mind the conditions I've underlined.

<u>If acceleration is not constant and changes over time</u>, so that the acceleration is some function of time a(t), then you can determine the velocity function v(t) by using the fundamental theorem of calculus. You need to know a particular velocity for some time to completely characterize v(t), though. For example, if you're given the initial velocity v[initial] = v(0), then

\displaystyle v(t) = v(0) + \int_0^t a(u) \, du

or if you know any other velocity for some time t₀ > 0,

\displaystyle v(t) = v(t_0) + \int_{t_0}^t a(u) \, du

8 0
3 years ago
A car with an initial speed of 4.30 m/s accelerates at a rate of 3 ms^2. What is the car's speed after 5 s?
Lorico [155]

Answer:

v = u + at

u = 4.30 m/s

a = 3 m/s^2

t = 5 s

v = 4.30 + (3)(5)

v = 4.30 + 15

v = 19.30 m/s

Explanation:

Hope this helps!

5 0
3 years ago
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