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Keith_Richards [23]
3 years ago
10

What type of power plant is virtually pollution free

Physics
1 answer:
soldi70 [24.7K]3 years ago
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.<br><br><br> True or False
eimsori [14]

Answer: False

Explanation:

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

In thise sense, one of the best examples of this type of wind are the <em>Westerlies</em>, 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:

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

<em>-- The greater the force applied, the greater the acceleration.</em>

For the <u><em>other </em></u>statements in the question:

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

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

-- <em>An object at rest or in motion will remain at rest or in motion unless acted upon by an unbalanced force. </em>  True; comes from Newton's <em><u>1st </u></em>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:

<u>Given:</u>

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
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