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kaheart [24]
3 years ago
12

Monitoring systems may use ____, which are devices that respond to a stimulus (such as heat, light, or pressure) and generate an

electrical signal that can be measured or interpreted.
Physics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

idk

Explanation:

Pay attention more kid

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max2010maxim [7]
Air pollution!! Can I have brainlyist:))
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2 years ago
Objects in space that are moving at a constant velocity in a straight line ___________.
nikdorinn [45]
The best answer is A) <span>keep moving at a constant velocity until some forces act on them

As the man you're probably tired of hearing about said:

"Every object persists in its state of rest or in uniform motion in a straight line unless a new force acts upon it" 
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Put more simply, objects in motion tend to stay in motion, and tend the maintain the same velocity (direction and speed) and objects at rest tend to stay at rest. </span>
6 0
3 years ago
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5 0
2 years ago
you throw an object straight up at 15m/s. how fast is it going when it gets back to the height from which you threw it?
laiz [17]
At the same speed because it will slow down as it approaches the peak then speed up as it goes down again


it will be going 15m/s when it gets to the same height if we neglect air resistance and the object doesn't hit something
8 0
2 years ago
One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista
Jlenok [28]

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where  s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

     g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g  is

g (planet) = 0.096g

5 0
3 years ago
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