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Goshia [24]
2 years ago
7

Give two examples of motion related to civil engineering.​

Physics
1 answer:
krok68 [10]2 years ago
7 0

rotational motion of a concrete mixer

linear motion of a tractor

is this what you mean

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cooling down after a workout allows the oxygen to continuing moving through the body and keeps the muscles from tightening up to
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I think the answer is true
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in which type of wave are vibrations at right angles to the direction in which the wave is travelling​
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longitudinal waves have those properties

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2 years ago
If Fg=mg solve for g
Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

6 0
3 years ago
Which instrument produces the sound of a single frequency?
notka56 [123]
NO musical instrument produces a 'pure' tone with only a
single frequency in it.

EVERY instrument produces more or less harmonics (multiples)
in addition to the basic frequency it's playing.

The percussion instruments (drums etc) are the richest producers
of bunches of different frequencies.

Fuzzy electric guitars are next richest.

The strings and brass instruments are moderate producers of
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Then come the woodwinds ... clarinet, oboe, etc.

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6 0
3 years ago
Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis
solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

v^{2}=u^2+2as

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equalsT=2\times 30.51\approx61seconds

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 39.66m/s

4 0
3 years ago
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