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goldfiish [28.3K]
2 years ago
7

Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or

not at all.
Physics
1 answer:
slamgirl [31]2 years ago
5 0

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

To learn more about feedback loop, visit:

brainly.com/question/11312580

#SPJ4

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An object of mass 6 kg. is resting on a horizontal surface. A horizontal force
son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force (W), measured in joules, is defined by the following expression:

W = F\cdot \Delta x (1)

Where:

F - Force, measured in newtons.

\Delta x - Distance, measured in meters.

If we know that F = 15\,N and \Delta x = 100\,m, then the work done by the force exerted on the object is:

W = (15\,N)\cdot (100\,m)

W = 1500\,J

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object (a), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2} (2)

Where v_{o} is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

\frac{1}{2}\cdot a \cdot t^{2} =  \Delta x-v_{o}\cdot t

a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}

If we know that \Delta x = 100\,m, v_{o} = 0\,\frac{m}{s} and t = 10\,s, then the net acceleration experimented by the object is:

a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}

a = 2\,\frac{m}{s^{2}}

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

\Sigma F = F - f = m\cdot a (3)

Where:

F - External force exerted on the object, measured in newtons.

f - Kinetic friction force, measured in newtons.

If we know that F = 15\,N, m = 6\,kg and a = 2\,\frac{m}{s^{2}}, the kinetic friction force is:

f = F-m\cdot a

f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)

f = 3\,N

The work done by friction (W'), measured in joules, is:

W' = f\cdot \Delta x (4)

W' = (3\,N) \cdot (100\,m)

W' = 300\,J

And the net work experimented by the object is:

\Delta W = 1500\,J - 300\,J

\Delta W = 1200\,J

By the Work-Energy Theorem we understand that change in translational kinetic energy (\Delta K), measured in joules, is equal to the change in net work. That is:

\Delta K = \Delta W (5)

If we know that \Delta W = 1200\,J, then the change in translational kinetic energy is:

\Delta K = 1200\,J

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0
3 years ago
In a basketball game, a player shoots a jump shot. Which force actually causes the player to jump? the player pushing down on th
Gemiola [76]

Answer:  the player pushing down on the floor

Explanation:

4 0
3 years ago
Read 2 more answers
I don’t know what to do help me
kirill115 [55]

Answer:

oooh thats hard

Explanation:

well i would probaly search the page number or anthing on the page that   should help

4 0
3 years ago
What total mass must be converted into energy
Eduardwww [97]

This question apparently wants you to get comfortable
with  E = m c² .  But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
                                                    E = m c²

                           1.03 x 10⁻¹³ joule  =  (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

                         Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

                                   =  (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

                                   =        1.144 x 10⁻³⁰  kg .    (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature.  The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:
 
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

             =  (2,242 x 10⁶ x 86,400 x 365) joules

             =          7.0704 x 10¹⁶ joules .

How much converted mass is that ?

                                           E  =  m c²

Divide each side by  c² :    Mass  =  E / c² .
c = 3 x 10⁸ m/s

              Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

                        =        0.786 kilogram ! ! !

THAT should impress us !  If I've done the arithmetic correctly,
then roughly  (1 pound  11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>
7 0
3 years ago
Read 2 more answers
SP: Calculate the moment
ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, F=80\ N

Distance of application of force from the point about which moment is needed is, d=25\ cm=\frac{25}{100}\ m=0.25\ m

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0
3 years ago
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