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Ghella [55]
2 years ago
7

Noise from highways is a large issue, particularly for those who live close to the highway. Engineers have come up with several

solutions to the problem. One solution can be a high vertical wall. Second, planting vegetation, like trees. Third, using a large buffer zone or open space.
Explain how trees, a high wall or buffer zones can reduce the noise from a highway that is heard at a nearby home.
Think…
· How would the noise be for someone in a home?
· How would the noise be for someone driving on the highway?
· How does the DOPPLER EFFECT fit in to this scenario?
You must write a response with a MINIMUM of 5 sentences to receive full credit.
WILL MARK AS BRAINILEST
Physics
1 answer:
kifflom [539]2 years ago
8 0
Noise from highways are a significant issue. particularly for those who live close to the highway
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Lawson's criterion states that the product of ion density and confinement time must exceed a certain number before a break-even
LekaFEV [45]

According to Lawson's criterion, the outcome is determined by the product of ion density and confinement time because the temperature must be maintained for a sufficient confinement time and with a sufficient ion thickness to obtain a net gain of power from a fusion reaction.

<h3>What are Lawson's criterion?</h3>
  • The overall conditions that must be met in order to produce more energy than is required for plasma heating are usually expressed in terms of the product of ion density and confinement time, a condition known as Lawson's criterion.
  • In nuclear fusion devices, confinement time is defined as the amount of time the plasma is kept at a temperature above the critical ignition temperature.
  • Even at temperatures high enough to overcome the coulomb barrier to nuclear fusion, a critical density of ions must be maintained in order to achieve a net yield of energy from the reaction.
  • Because the density required for a net energy yield is correlated with the confinement time for hot plasma, the minimum condition for a productive fusion reaction is typically stated in terms of the product of ion density and confinement time, which is known as Lawson's criterion.

To learn more about Lawson's criterion, refer:

brainly.com/question/28303495

#SPJ4

7 0
1 year ago
Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

K =1/4 \pi \epsilon_0 = 8.99 \times 10^9 N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

E_1 + E_3 = 0

E =E_2 + E_4

E = 2 E_2

[E_2 =\frac{2\times k \times q}{r^2}

[r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m]

plugging all value

E = 2 E_2

E = 2 E_2 =\frac{2\times k \times q}{r^2}

E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}

E = 440816.32 N/C

3 0
3 years ago
Read 2 more answers
In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g
Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η=\frac{electric power}{water power}

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

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Vikki [24]

Answer:

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What is the difference between first aid and first aid kit
Archy [21]

Answer:

One has the word kit and the other doesn't

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