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pashok25 [27]
3 years ago
14

What does the term equilibrium refer to?

Physics
1 answer:
snow_lady [41]3 years ago
5 0
<h3><u>Answer</u>;</h3>

A) the resting position of the wave

<h3><u>Explanation</u>;</h3>
  • A wave is a transmission of a disturbance from one point which is the source to another, and this involves transfer of energy through a material medium.
  • <em><u>Equilibrium refers to a state of balance between opposing forces, it is a state of balance in which opposing forces cancel one another. </u></em>
  • <em><u>When wave is in rest position its called equilibrium position of a wave. When a wave travels through a material medium, the particles in the medium are disturbed from their resting, or equilibrium positions.</u></em>
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Select the correct answer.
Serhud [2]

Answer:

b hope this helps

Explanation:

5 0
3 years ago
Read 2 more answers
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
57:07
Reptile [31]

Answer:

v (speed) = S / t = 4 * 400 m / (6 * 60 sec) = 4.4 m/s

The average velocity  is zero because there is no net vector displacement.

5 0
3 years ago
Which type of wave has a frequency just higher than violet visible light and is used by the skin to produce Vitamin D?
ANTONII [103]
Ultraviolet rays can help us get vitamin D, but they can also damage or skin if w are exposed for too long.

Your answer is D.
4 0
3 years ago
190 kg of water is to be raised by a water pump to a height of 25 meters from the bottom of a well in 60 seconds. What should be
ladessa [460]

Answer:

776.6 w

1.04 hp

Explanation:

given:

Mass, m = 190kg

height change, h = 25m

time elapsed, t = 60 s

acceleration due to gravity, g = 9.81 m/s²

Potential energy required raising 190 kg of water to a height of 25m

= mgh

= 190 x 9.81 x 25

= 46,597.5 J

Power required in 60 s

= Energy required ÷ time elapsed

= 46,597.5 ÷ 60

= 776.6 Watts  (Use conversion 1 W = 0.00134102 hp)

= 776.6 w x 0.00134102 hp/w

= 1.04 hp

6 0
3 years ago
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