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KIM [24]
3 years ago
11

Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e

xploration, or a combination of sources? Are there other efforts that should be explored?
Physics
2 answers:
True [87]3 years ago
7 0
The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.
telo118 [61]3 years ago
5 0

Answer:

I believe that the government should focus on supporting solar and wind energy over anything else.

Explanation:

There are two reasons why I believe the government should put the greatest efforts in solar and wind energy.

First of all, obtaining coal, natural gas, and oil is cheap because the governments and companies have put great amounts of money in research to produce it at the lowest possible costs. Therefore if we fund research with the same model, we are going to obtain cheaper costs. The second reason is that they are renewable and 0 pollution sources of energy, unlike coal, gas, and il. Also, we can create engines like the ones electric cars use, to all transports.

There are other efforts like nitrogen, geothermic energy and my favorites, gravitational rejection impulse energy and electromagnetic energy like the one the Japanese shinkansen uses.

You might be interested in
The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason
Tamiku [17]

The MCB of a rupas room is tripped and keeps on tripping again and again, and if it is a domestic circuit, what could be the reason for this phenomenon?

The reason could be a short circuit which is resulting in higher level of currents to pass through the MCB which is resulting in trip every time.

OR

The MCB is faulty and might need a replacement.

To Diagnose the problem further more.

Turn off all the switches in rupas room and then try turning on the MCB. If it trips again then MCB is faulty (Subjective to the fact there everything was normal before this issue and no signs of short circuit or spark in wiring were observed)

If MCB does not trip in point 1 then Turn ON all the switches one by one. This shall give you the cause of problem.

4 0
3 years ago
A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
Marysya12 [62]

Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

6 0
3 years ago
This back-and-forth movement of electrons is called . In contrast, the movement of electrons in one direction in a battery circu
stich3 [128]
The back-and-forth movement of electrons is called alternating current. Electrons go back and forth, the direction of their path alternates from one direction to another.

the movement of electrons in one direction is called direct current. The electrons move in a direct, single path without changing directions.
5 0
3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

solving for x we get

\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

x = 0.200 m

b)electric field at half way mean x =0.25

E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}

E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

Mass of object = 2.00 kg

Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

W=F\cdot d

W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0
3 years ago
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