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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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The radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm?
marin [14]

Using <span>r </span> to represent the radius and <span>t </span> for time, you can write the first rate as:

<span><span><span><span>dr</span><span>dt</span></span>=4<span>mms</span></span> </span>

or

<span><span>r=r<span>(t)</span>=4t</span> </span>

The formula for a solid sphere's volume is:

<span><span>V=V<span>(r)</span>=<span>43</span>π<span>r3</span></span> </span>

When you take the derivative of both sides with respect to time...

<span><span><span><span>dV</span><span>dt</span></span>=<span>43</span>π<span>(3<span>r2</span>)</span><span>(<span><span>dr</span><span>dt</span></span>)</span></span> </span>

...remember the Chain Rule for implicit differentiation. The general format for this is:

<span><span><span><span><span>dV<span>(r)</span></span><span>dt</span></span>=<span><span>dV<span>(r)</span></span><span>dr<span>(t)</span></span></span>⋅<span><span>dr<span>(t)</span></span><span>dt</span></span></span> </span>with <span><span>V=V<span>(r)</span></span> </span> and <span><span>r=r<span>(t)</span></span> </span>.</span>

So, when you take the derivative of the volume, it is with respect to its variable <span>r </span> <span><span>(<span><span>dV<span>(r)</span></span><span>dr<span>(t)</span></span></span>)</span> </span>, but we want to do it with respect to <span>t </span> <span><span>(<span><span>dV<span>(r)</span></span><span>dt</span></span>)</span> </span>. Since <span><span>r=r<span>(t)</span></span> </span> and <span><span>r<span>(t)</span></span> </span> is implicitly a function of <span>t </span>, to make the equality work, you have to multiply by the derivative of the function <span><span>r<span>(t)</span></span> </span> with respect to <span>t </span> <span><span>(<span><span>dr<span>(t)</span></span><span>dt</span></span>)</span> </span>as well. That way, you're taking a derivative along a chain of functions, so to speak (<span><span>V→r→t</span> </span>).

Now what you can do is simply plug in what <span>r </span> is (note you were given diameter) and what <span><span><span>dr</span><span>dt</span></span> </span> is, because <span><span><span>dV</span><span>dt</span></span> </span> describes the rate of change of the volume over time, of a sphere.

<span><span><span><span><span>dV</span><span>dt</span></span>=<span>43</span>π<span>(3<span><span>(20mm)</span>2</span>)</span><span>(4<span>mms</span>)</span></span> </span><span><span>=6400π<span><span>mm3</span>s</span></span> </span></span>

Since time just increases, and the radius increases as a function of time, and the volume increases as a function of a constant times the radius cubed, the volume increases faster than the radius increases, so we can't just say the two rates are the same.

7 0
2 years ago
An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How
Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

g=24 m/s^2

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

s=0+\frac{1}{2}(24)(1)^2=12 m

6 0
3 years ago
PLZ ANSWER ASAP IF YOU KNOW THIS...At the county fair, Carrie and Sam climbed up on the carousel horses. Around and around they
Maslowich
The answer is (a) because movement is acceleration 
8 0
2 years ago
Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe
dusya [7]

Answer:

Acceleration of the ship, a=2.14\times 10^{-7}\ m/s^2

Explanation:

It is given that,

Mass of both ships, m=39000\ metric\ tons=39\times 10^6\ kg

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

F=G\dfrac{m^2}{d^2}

F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

a=\dfrac{F}{m}

a=\dfrac{8.38\ N}{39\times 10^6\ kg}

a=2.14\times 10^{-7}\ m/s^2

So, the acceleration of either ship due to the gravitational attraction of the other is 2.14\times 10^{-7}\ m/s^2. Hence, this is the required solution.

7 0
3 years ago
A snail and an inchworm are in a race. Their race track heads north for a distance of 2 m. If the inchworm comes to the end of t
olga2289 [7]
Well the basic equation for velocity is v=d/t where d is distance and t is time. So v=2m/50s and the answer is v=0.04meter/second.
7 0
3 years ago
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