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2 years ago

Select three functions of catalysts.

Engineering

1 answer:

Korolek [52]2 years ago

5 0

Answer: speed up food processing

speed up plant growth

Increase fuel efficiency

Explanation:

A catalyst simply refers to a substance that leads to an increase in the reaction rate when it's added to a substance. When the activation energy is reduced by catalysts, this.hwlpa on the speeding up of a reaction.

Therefore,the functions of catalysts include speed up food processing, speeding up plant growth and increase fuel efficiency

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Ordan has _ 5 8 can of green paint and _ 3 6 can of blue paint. If the cans are the same size, does Jordan have more green paint

Morgarella [4.7K]

Answer:

Jordan has more green paints

Explanation:

Given

$Green = \frac{5}{8}$

$Blue = \frac{3}{6}$

Required

Which paint does he have more?

For better understanding, it's better to convert both measurements to decimal.

For the green paint:

$Green = \frac{5}{8}$

$Green = 0.625$

For the blue paint:

$Blue = \frac{3}{6}$

$Blue = 0.5$

By comparison:

$0.625 > 0.5$

<em>This means that Jordan has more green paints</em>

3 0

3 years ago

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

Arlecino [84]

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

8 0

3 years ago

I NEED HELP HERE SUM POINTS ( i will report if you put a link and or no answer

Y_Kistochka [10]

Answer:

ok I will help you ha Ha ha ha ha ha ha ha ha ha ha ha

8 0

2 years ago

True or False: Stress can effectively be relieved through physical activity, getting enough rest and sleep, and relaxation techn

Neko [114]

Answer:

True

Explanation:

Actually this are some of the nitty gritty answers and ways to control or stip stress, lemme explain them you see stress most of the time may come from deep thoughts that are and are pushing you to the wall, and you in your state you as you react to that which is so demanding as it makes you tense,so if at you will incorporate physical activities like running, jogging or push ups the psychological tension in you is broken as you focus so much on the activities rather than the tension, then getting enough rest cools the mind and all your thoughts settle as in your brain starts to adapt to chilling and relaxation and enough sleep will actually make ones head to be at peace because if you lack enough sleep you might have an excruciating migraine when you are stressed up and finally relaxation techniques makes the body accept the situation and then you manouver out of it as you grow strongly.

Hope this will help!

7 0

2 years ago

Read 2 more answers

For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or

Alborosie

Answer:

Glass: Low-Loss dielectric

α = 8.42*10^-11 Np/m

β = 468.3 rad/m

λ = 1.34 cm

up = 1.34*10^8 m/s

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = 9.75 Np/m

β = 12.16 rad/m

λ = 51.69 cm

up = 0.52*10^8 m/s

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = 6.3*10^-4 Np/m

β = 6.3*10^-4 Np/m

λ = 10 km

up = 0.1*10^8 m/s

ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to determine category of the medium as follows:

σ / w*εr*εo

Where, w is the angular speed of wave

εo is the permittivity of free space = 10^-9 / 36*pi

- Now we classify as follows:

$Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\$

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.

Glass: Low-Loss dielectric

Tissue: Quasi-Conductor

Wood: Good conductor

- Now we will use categorized material base equations from Table 17-1 as follows:

Glass: Low-Loss dielectric

α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

α = 8.42*10^-11 Np/m

β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

β = 468.3 rad/m

λ = 2*pi / β = 2*pi / 468.3

λ = 1.34 cm

up = λ*f = 0.0134*10^10

up = 1.34*10^8 m/s

ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

α = 9.75 Np/m

β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

β = 12.16 rad/m

λ = 2*pi / β = 2*pi / 12.16

λ = 51.69 cm

up = λ*f = 0.5169*100*10^6

up = 0.52*10^8 m/s

ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

β = α = 6.3*10^-4 Np/m

λ = 2*pi / β = 2*pi / 6.3*10^-4

λ = 10 km

up = λ*f = 10,000*1*10^3

up = 0.1*10^8 m/s

ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

ηc = 6.28*( 1 + j )

8 0

3 years ago