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2 years ago

The flow control circuit design that can best control an overrunning load is the ____ circuit.

Physics

1 answer:

lara31 [8.8K]2 years ago

3 0

The meter out circuit is the flow control circuit design that can most effectively control an overrunning load.

The meter-out circuit can be very accurate, but are not efficient. The meter-out circuit can control overrunning as well as opposing loads while the other one method must be used with opposing loads only. The choice of flown control valve method and the location of the flow control in the circuit are dependent on the type of application being controlled.

<h3>What is a Circuit ?</h3>

In electronics, a circuit is a complete circular conduit through which electricity flows. A simple circuit consists of conductors, a load, and a current source. The term "circuit" broadly refers to any continuous path via which electricity, data, or a signal might flow.

The directional valve shifts, causing the actuator to move faster than pump flow can fill it due to an overrunning load. Oil is leaking from one side, whereas there is none on the other.

Hence, flow control circuit design that can best control an overrunning load is the opposing circuit

Learn more about Circuit here:

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Mention three features of the Constitution and a note about them

Dvinal [7]

Answer:

Explanation:(1) a constitution is a supreme law of the land, (2) a constitution is a framework for government; (3) a constitution is a legitimate way to grant and limit powers of government officials. Constitutional law is distinguished from statutory law.

7 0

3 years ago

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the

fredd [130]

Answer:

t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material = 1.375

we know the equation

c = ν λ

where ν is the frequency of the wave

c is the speed of light

$\nu= \dfrac{c}{\nu\lambda}$

$\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}$

ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

$t = \dfrac{\lambda}{4\mu_{material}}$

$t = \dfrac{522 \times 10^{-9}}{4\times 1.375}$

t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

7 0

3 years ago

Which colors allowed the most electricity to be generated

Naddik [55]

Yellow and red hope that helped

3 0

3 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago