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White raven [17]
2 years ago
14

When you sweat, what is the external stimuli? I need help asap. ​

Physics
1 answer:
mixer [17]2 years ago
8 0

Answer:

An External Stimulus is a stimulus that comes from outside an organism. Examples: You feel cold so you put on a jacket. When you sweat, the external stimulus is either you're anxious or hot.

Explanation:

hope it helps! <3

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PLS HELP DUE IN 30 MINS!!! True or false: A large-amplitude pulse travels at a faster speed than a small-amplitude pulse. Explai
OLga [1]

Answer:

False

Explanation:

Larger pulse with more energy means that the wave has large amplitude. As in the given question, both the waves are travelling in the same medium, hence their speeds will remain same and therefore the larger pulse will not overtake the smaller pulse. Remember the amplitude of a wave does not affect the speed at which the wave travels

8 0
2 years ago
Read 2 more answers
What is the drawback to using superconductors?
Fittoniya [83]

Answer:

Option A

The cost of keeping the semiconductor below the critical temperature is unreasonable

Explanation:

First of all, we need to understand what superconductors are. Superconductors are special materials that conduct electrical current with almost zero resistance. This means that there is little or no need for a voltage source to be connected to them. As a matter of fact, once a superconductor is connected to a power supply, one can remove the power supply and the current will still flow.

However, most superconducts can only conduct at very low temperatures up to -200 degrees Celcius. This is because, at that temperature, their atoms and molecules are relatively settled, hence they pose little or no resistance to the flow of current.

This as you can guess is extremely difficult to do, as you will need a lot of effort to cool it to that temperature and maintain it.

This makes option a the answer:

The cost of keeping the semiconductor below the critical temperature is unreasonable.

7 0
2 years ago
A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go? (b) how long is it in the air
Mariulka [41]
Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
where
h = vertical height, m
g = 9.8 m/s², acceleration due to gravity
t = time, s

To find how long the ball stays in the air, set h = 0 to obtain
4.9t² - 22t = 0
t(4.9t - 22) = 0
t = 0, or t = 22/4.9 = 4.49 s
t = 0 corresponds to the launch, and t = 4.49 s corresponds to when the ball retuns to the ground.

Answer: The ball stays in the air for 4.5 s (nearest tenth)

3 0
3 years ago
Which lists types of materials from most conductive to least conductive?
Lilit [14]
Superconductor, conductor, semiconductor, insulator
7 0
3 years ago
Read 2 more answers
Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes
pantera1 [17]

Answer:

The  magnitude of F1 is

|F1|=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.  

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.  

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

\displaystyle -2F\ sin21^o+F\ cos\alpha =0

The sum of the vertical components of F1 and F2 must equal the total force Ft

\displaystyle -2F\ cos21^o+F\ sin\alpha =460

Solving for \alpha in the first equation

\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o

\displaystyle cos\alpha =0.717=>\alpha =44.214^o

\displaystyle F(2cos21^o+sin\alpha)=460

\displaystyle F=\frac{460}{2cos21^o+sin\alpha}

\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}

\displaystyle F=179.37\ N

The  magnitude of F1 is

|F1|=2*F=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

4 0
3 years ago
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