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2 years ago

What parameters we control through a valve?

Physics

1 answer:

liq [111]2 years ago

8 0

Answer:

Control valves are used in many processes to control flow, pressure, temperature or other variables. The type of valve used will depend on the size of the pipe, the overall pressure that the system operates, the flowing media, process conditions, and other factors.

Also

A control valve is a valve used to control fluid flow by varying the size of the flow passage as directed by a signal from a controller. This enables the direct control of flow rate and the consequential control of process quantities such as pressure, temperature, and liquid level.

Hope this helps :)

Pls brainliest...

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Match each term to its definition.

den301095 [7]

Answer: first one is electrochemical
Second one is combustion
Third one is photosynthesis
Fourth one is respiration

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3 years ago

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Which planet has the most extreme temperature variations

vichka [17]

Answer:

I believe Mercury has the most extreme temperatures in the solar system, ranging from -280?F at night to 800 degrees F during the day for parts of the surface.

Hope that helps! :)

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3 years ago

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The current in a lamp is 0.5 ampere when it is plugged into a standard wall outlet. What is the resistance of the lamp when it i

lidiya [134]

The resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms)

Explanation:

In the United States Of America the standard voltage is 120 v and their frequency is 60 Hz

Standard wall outlet voltage is 120 V

The current in the lamp is 0.5 ampere

Resistance (R) = V/ I

= 120/0.5

= 240Ω (ohms)

Thus the resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms).

8 0

3 years ago

A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her

AlekseyPX

Answer:

28.025 Nm

Explanation:

Angular acceleration, α = 29.5 rad/s^2

oment of inertia, I = 0.95 kg m^2

The torque is defined as

τ = I x α

τ = 0.95 x 29.5

τ = 28.025 Nm

Thus, the torque is 28.025 Nm.

Explanation:

7 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago