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Gelneren [198K]
3 years ago
8

Which choice shows the correct sequence of features formed by continued wave erosion?

Physics
1 answer:
attashe74 [19]3 years ago
7 0

Answer:

Wave-cut cliff, sea arch, sea stacks

Explanation:

The effect of a wave erosion is made obvious by the structures formed by the wave action.

The high areas of land adjacent to the incoming wave develop the early features or the formation of wave action, which includes the <em>wave-cut cliff</em>

The continuous undercutting of the cliff by the wave results in the formation of the <em>wave cut platform</em>

The effect of the wave further on a cliff, results in the formation of a sea arch and finally a <em>sea stack</em>

Therefore, the correct sequence is the wave-cut cliff, sea arch, sea stacks

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Almost everything in the universe orbits around a central object. The planets orbit the sun and moons orbit planets. The force M
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Gravity is why orbits are happening.
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2. What is the resistance expected on a heater that is feed by 480V AC and the Electric Power is 4.6kW? ​
Korvikt [17]

The resistance expected of the heater is 50.1 ohms.

<h3>What is resistance?</h3>

Resistance can be defined as the opposition to the flow of electric current in an electric circuit. The S.I unit of resistance is Ohms (Ω).

To calculate the resistance of the heater, we use the formula below.

<h3>Formula:</h3>
  • R = V²/P............. Equation 1

Where:

  • R = Resistance of the heater
  • P = Power of the heater
  • V =  Voltage supplied to the heater

From the question,

Given:

  • V = 480 V
  • P = 4.6 kW = 4600 W

Substitute these values into equation 1

  • R = (480²)/4600
  • R = 50.1 ohms.

Hence, the resistance expected of the heater is 50.1 ohms.

Learn more about resistance here: brainly.com/question/17563681

3 0
2 years ago
Definition of Speed?​
ANEK [815]

Explanation:

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2 years ago
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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
Một người đi đều với vận tốc 1,5 m/s, muốn đi quãng đường dài 1,5 km thì người
andrezito [222]
T=s/v=>t=1500/1,5=1000s
1,5km=1500m
6 0
2 years ago
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