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3 years ago

What type of power plant is virtually pollution free

1 answer:

6 0

A power plant that is virtually pollution free would be a power plant that does not consume gas, petrol, coal or any of the other resources that may cause pollution.

The virtually pollution-free plant will be mainly depending on natural renewable sources of energy

Examples:
1- geothermal plants (including dry steam plants, binary cycle plants and flash steam plants)
2- solar plant
3- wind turbines
4- water turbines

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Winds blowing toward the east are called easterlies. True or False

eimsori [14]

Answer: False

Explanation:

Winds are named for the cardinal direction they blow from. Hence, a wind that "blows towards the east", logically should come from the west and is called a "west wind".

In thise sense, one of the best examples of this type of wind are the Westerlies, which are are prevailing winds that blow from the west at midlatitudes and have the characteristic that are stronger during winter and weaker during summer.

Therefore, the statement is false.

6 0

3 years ago

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0

3 years ago

What is Newton's Second Law of Motion? (2 points) Every reaction is equal to the force applied. Forces are balanced when they ar

raketka [301]

Well, they're not quite the way Newton expressed it, but out of all this mess of statements, there are two that are correct AND come from Newton's 2nd Law of Motion:

-- The smaller the mass of an object, the greater the acceleration of that object when a force is applied.

-- The greater the force applied, the greater the acceleration.

For the other statements in the question:

-- Every reaction is equal to the force applied. True; comes from Newton's 3rd law of motion.

-- Forces are balanced when they are equal and opposite. True; kind of a definition, not from Newton's laws of motion.

-- An object at rest or in motion will remain at rest or in motion unless acted upon by an unbalanced force. True; comes from Newton's 1st law of motion.

5 0

3 years ago

Each step in any energy conversion process will _____. gain energy create energy dissipate energy destroy energy

Damm [24]

The correct answer is to dissipate energy. Each step in any energy conversion process will dissipate energy. The energy conversion is the process in which a form of energy is being converted into a new or another form of energy.

7 0

3 years ago

Read 2 more answers

In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top pla

lianna [129]

Answer: See below

Explanation:

Given:

The potential between plates, V = 240 V

Distance between plates, d = 0.02 m

The mass of drop, m = 2x10^-11

Charge on electron, e = 1.6x10^-19

Part (a)

The free-body diagram is attached below

Part (b)

The electric field is given by,

$E=\frac{V}{d}$

On applying force balance, the force on oil drop is equal to the weight of the oil,

$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$

Substituting the given values in the above equation,

$\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}$

Therefore, the charge on the oil drop is 1.63x10^-14 C

Part (c)

There will be an excess of electrons on the oil drop.

The number of electrons on oil drop can be calculated as,

$\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}$

Therefore, the number of excess electrons is 1.01x10^5

3 0

1 year ago