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Which of the Following Strategies Can Help Earth's Coal Supply Last Longer

Written By Lewis Thaversonly Thursday, February 17, 2022 Add Comment Edit

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Which of the post-obit strategies can help Globe'due south coal supply last longer?

ii answers:

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D. Teach the public free energy conservation

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Answer: D. Teach the public free energy conservation

Coal is an efficient and cheap source of energy which provides a stable and large scale free energy generation in the form of electrical and thermal energy. Information technology is a type of fossil fuel which is virtually abundantly nowadays. But it is also a non- renewable source of energy, and cannot be replenished afterwards a single utilize. It is used as a source of generation of electricity, production of metallic objects, manufacturing of a liquid fuel, engines of local trains and for other purposes. Equally, it is a non- renewable resource it should exist used in a sustainable mode.

Hence, D. Teach the public energy conservation is the correct option.

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I think the right reply is A mark has greater gravitational potential energy
:)

d = distance between the two point charges = 60 cm = 0.lx m

r = distance of the location of point "a" where the electric field is zero from charge $q_{1}$ between the 2 charges.

$q_{1}$ = magnitude of charge on 1 accuse

$q_{2}$ = magnitude of charge on other accuse

$q_{1}$ = 3 $q_{2}$

$E_{1}$ = Electrical field by accuse $q_{1}$ at betoken "a"

$E_{2}$ = Electrical field by charge $q_{2}$ at signal "a"

Electric field by charge $q_{1}$ at bespeak "a" is given as

$E_{1}$ = g $q_{1}$ /r²

Electrical field by charge $q_{2}$ at bespeak "a" is given every bit

$E_{2}$ = k $q_{2}$ /(d-r)²

For the electrical field to be zero at point "a"

$E_{2}$ = $E_{1}$

k $q_{2}$ /(d-r)² = yard $q_{1}$ /r²

$q_{2}$ /(d-r)² = 3 $q_{2}$ /r²

ane/(0.60 - r)² = 3 /r²

r = 0.38 yard

r = 38 cm

The correct respond is B)<span>graduated cylinder, beaker, thermometer and hot plate.
</span>Cylinder to measure out information technology, chalice to contain information technology, Thermometer to mensurate the temperature alter and hot plate to heat the water.

(a) $a = -\mu_k g$

In gild to find the acceleration of the puck, we must clarify the forces acting on it.

Forth the vertical management, nosotros have 2 forces: the weight of the puck, mg, and the normal reaction, Due north. The two forces are equal and reverse and then that the vertical acceleration is naught, and so nosotros can write

$N-mg = 0\rightarrow N = mg$ (1)

where one thousand is the mass of the puck and g is the acceleration of gravity.

Along the horizontal direction, at that place is simply one force interim on the puck: the strength of friction, opposite to the direction of motility of the puck. So we can write:

$-\mu_k N = ma$ (2)

where $\mu_k$ is the coefficient of kinetic friction and a is the acceleration. Substituting (Northward) from eq.(one) into (ii), we discover an expression for a:

$-\mu_k mg = ma\\a = -\mu_k g$

(b) $d=\frac{v_0^2}{2\mu_kg}$

We can find the distance that the puck slides using the post-obit SUVAT equation:

$v^2-v_0^2 = 2ad$

where

v = 0 is the concluding velocity of the puck (it comes to a terminate, so v = 0)

$v_0$ is the initial velocity of the puck

a is the acceleration

d is the distance travelled

Substituting the expression for (a) that we plant previously, we find an expression for d:

$d=\frac{-v_0^2}{2a}=\frac{-v_0^2}{-2\mu_k g}=\frac{v_0^2}{2\mu_kg}$

Answer:

<h3>ii.3125m/s²</h3>

Explanation:

Using the equation of motion v² = u²+2aS

5 is the last velocity = 120km/hr

120km/hr = 120 * 1000/i * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

South is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = two.3125m/south²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 thou is 2.3125m/south²

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Source: https://en-ya.guru/physics/question-1246514.html

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