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1 year ago

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

Physics

1 answer:

lara31 [8.8K]1 year 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:

brainly.com/question/26064065

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A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati

Ainat [17]

Answer:

The asteroid's acceleration at this point is $2.71\ m/s^2$

Explanation:

The equation that governs the trajectory of asteroid is given by :

$x=6.5t-2.3t^3$

The velocity of asteroid is given by :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2$

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

$6.5-6.9t^2=0\\\\t=0.971\ s$

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t$

Put t = 0.971 s

$a=-13.8\times 0.197\\\\a=-2.71\ m/s^2$

So, the asteroid's acceleration at this point is $2.71\ m/s^2$ and it is decelerating.

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3 years ago

In outer space, a piece of rock continues moving at the same velocity for

Ray Of Light [21]

The absence of external force in the outer space, allows the piece of rock to continue moving at the same velocity for thousands of years.

<h3>Absence of external force on the outer space</h3>

The outer space is almost an absolute vacuum, because it's nearly empty. There is no matter such as air in the outer space that will provide an external force needed to change the velocity of the piece of rock.

From Newton's first law of motion, an object in a state of rest or uniform motion in a straight line, will continue in that state unless it is acted upon by an external force.

Thus, the absence of external force in the outer space, allows the piece of rock to continue moving at the same velocity for thousands of years.

Learn more about outer space here: brainly.com/question/24701339

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1 year ago

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed

olasank [31]

Answer:

Average acceleration is $(11.05)g\ m/s^2$

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{v}{t}$

$a=\dfrac{6500\ m/s}{60}$

$a=108.33\ m/s^2$

Since, $g=9.8\ m/s^2$

So, $a=(11.05)g\ m/s^2$

So, the angular acceleration of the missile is $(11.05)g\ m/s^2$ . Hence, this is the required solution.

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2 years ago

Lab: Thermal Energy Transfer What is the purpose of the lab, the importance of the topic, and the question you are trying to ans

VladimirAG [237]

Answer:

it's important because it shows how thermal energy transforms or continues to be all around us in everything

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2 years ago

A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m

ArbitrLikvidat [17]

Answer:

Explanation:

The sum of force along both components x and y is expressed as;

$\sum Fx = ma_x \ and \ \sum Fy = ma_y$

The magnitude of the net force which is also known as the resultant will be expressed as $R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}$

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

$a_x = \frac{d^2 x }{dt^2}$

$a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4 )\\\\a_x = \frac{d}{dt}(12t )\\\\a_x = 12m/s^{2}$

Similarly,

$a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2} )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2$

$\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N$

$R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N$

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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2 years ago