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Mandarinka [93]
3 years ago
14

1. This heating system maintains room temperature at or near a particular value, known as the .

Physics
1 answer:
stiv31 [10]3 years ago
5 0

Answer:

Explanation:

1. This heating system maintains room temperature at or near a particular value, known as the set point.

A temperature setpoint is the level at which the body attempts to maintain its temperature. When the setpoint is raised, the result is a fever.

2. You open the window, and a blast of icy air enters the room. The temperature drops to 17 degrees Celsius, which acts as a STIMULUS to the heating system.

3. The thermostat is a SENSOR that detects the stimulus and triggers a response.

Thermostat is use to turn off or on a switch, when the temperature is high or low

4. The heater turns on, and the temperature in the room INCREASE until it returns to the original setting.

This is the work of the thermostat above, when the temperature of the room is below a certain temperature the thermostat triggered the switch and keep increasing the temperature of the room until normal setting.

5. The response of the heating system reduces the stimulus. This is an example of NEGATIVE feedback.

Body temperature is regulated by negative feedback. The stimulus is when the body temperature exceeds 37 degrees Celsius, the sensors are the nerve cells with endings in the skin and brain, the control is the temperature regulatory center in the brain, and the effector is the sweat glands throughout the body.

6. The way this heating system maintains a stable room temperature is similar to the way an animal's body controls many aspects of its internal environment. The maintenance of a relatively constant internal environment is known as HOMEOSTASIS.

Humans rely on homeostasis to keep their core temperature hovering around 98.6 degrees Fahrenheit, so that their bodies can maintain proper function. Homeostasis is the ability to maintain a relatively stable internal state that persists despite changes in the world outside.

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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
You have a grindstone (a disk) that is 95.2 kg, has a 0.399 m radius, and is turning at 93 rpm, and you press a steel axe agains
olya-2409 [2.1K]

Answer:

angular acceleration is -0.2063  rad/s²

Explanation:

Given data

mass m = 95.2 kg

radius r = 0.399 m

turning ω = 93 rpm

radial force N  = 19.6 N

kinetic coefficient of friction  μ = 0.2

to find out

angular acceleration

solution

we know frictional force that is = radial force × kinetic coefficient of friction

frictional force = 19.6 × 0.2

frictional force = 3.92 N

and

we know moment of inertia  that is

γ =  I ×α = frictional force × r

so

γ  = 1/2 mr²α

α  = -2f /mr

α  = -2(3.92) /95.2 (0.399)

α  = - 7.84 / 37.9848 = -0.2063

so angular acceleration is -0.2063  rad/s²

3 0
3 years ago
Kinetic friction acts on a baseball player sliding into first base. Will the player's velocity change?
Law Incorporation [45]
Yes, his velocity will decrease the further he slides.
7 0
3 years ago
Force has _____. magnitude direction gravity weight
Contact [7]

Answer:

magnitude

Explanation:

7 0
3 years ago
Read 2 more answers
A mass weighing pounds is attached to a spring whose constant is lb/ft. The medium offers a damping force that is numerically eq
andreyandreev [35.5K]

Answer:

hello your question has some missing values attached below is the complete question with the missing values

answer :

a) 0.083 secs

b) 0.33 secs

c)  3e^-4/3

Explanation:

Given that

g = 32 ft/s^2 ,  spring constant ( k ) = 2 Ib/ft

initial displacement = 1 ft above equilibrium

mass = weight / g = 4/32 = 1/8

damping force = instanteous velocity  hence  β = 1

a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>

time mass passes through equilibrium = 1/12 seconds = 0.083

<u>b) Calculate the time at which the mass attains its extreme displacement </u>

time when mass attains extreme displacement = 1/3 seconds = 0.33 secs

<u>c) What is the position of the mass at this instant</u>

position = 3e^-4/3

attached below is the detailed solution to the given problem

6 0
2 years ago
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