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zhannawk [14.2K]
3 years ago
13

Which part of a feedback mechanism is able to monitor the conditions outside of cells and usually uses nerve cells to relay this

information to an integrating center?
A.) sensor
B.) effector
C.) stimulus
D.) response
Physics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer: A.) sensor

Explanation:

Homeostasis is the ability of the organism's inner body to regulate the internal environment in stable state with respect to the changes occurring in the external environment. It is usually done by the feedback controls.  

The maintenance of homeostasis within the body is essential. The following are the factors which controls the homeostasis. These includes:

1. Stimulus: It generate a response. It is an external factor which brings change in the internal body of the organism.

2. Receptor/ sensor: It detects the external stimulus and responds to the change.

3. Control center: The information from the receptor travels along the afferent pathway towards the control center. The function of the control center is to determine the response and controls the action.

4. Effector: The information from the control center travels down the efferent pathway to the effector. The function of the effector is to balance the stimulus to regulate and maintain homeostasis.

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A ship's boiler steam comes out at 112 C and pushes through the system, exiting into the condenser,
solniwko [45]

Answer:

D: 0.239

Explanation:

Equation for ideal efficiency is;

η = 1 - (T_c/T_h)

We are told that;

steam comes out at 112° C. Thus, T_h = 112°C. Converting to Kelvin gives; T_h = 112 + 273 = 385 K

The one exiting into the condenser is kept at 20°C. Thus; T_c = 20 + 273 = 293 K

Thus;

η = 1 - (293/385)

η = 0.239

5 0
3 years ago
The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo
Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

y(t) = h -\frac{1}{2}gt^2

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

x=v_x t

where

v_x is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

v_x = 18.5 m/s

we need to find the vertical velocity. This can be done by using the equation

v_y = u_y -gt

where

u_y =0 is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

v_y = -(9.8)(3.03)=-29.7 m/s

So the magnitude of the velocity at the impact (so, the speed at the impact) is

v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s

The angle instead can be found as:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}

so, 58.1 degrees below the horizontal.

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Neutral reaction is the reaction that occurs when an acid cancels with a base
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