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zavuch27 [327]
3 years ago
8

A student environmental group is creating a campaign for locally sourced energy resources.What would be best for them to feature

in their campaign?a coal mine located in their countya farmer with a field of solar panelsa company building a gas pipeline through the towna rancher drilling for oil in a horse pasture
Physics
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

<em>A farmer with a field of solar panels.</em>

Explanation:

The closest to a locally sources energy would have been

A coal mine located in their county.

But coal as an energy source is not environmentally friendly due to carbon emission, and should not be what the group should advocate for.

<em>The best bet for them is </em>

<em>A farmer with a field of solar panels.</em>

As solar panels are a source of green energy and green energy is what the environmental group should often and always advocate for

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6 0
3 years ago
Help ASAP PLEASE.............:3​
Evgesh-ka [11]

Answer:

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8 0
3 years ago
Find the acceleration of a train whose speed increases from 7m/s to 17 m/s in 120s
Ksenya-84 [330]
Initial speed of the train = 7 m/s
Final speed of the train = 17 m/s
Change of speed of the train = (17 - 7) m/s
                                               = 10 m/s
Time taken for the change of the speed of the train = 120 s
Then
Acceleration of the train = Change of speed of the train/Time taken for the change of speed
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                                       = 1/12 m/s^2
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7 0
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A spring loaded toy shoots straight upward with a velocity of 4.5 m/s. Determine the maximum height it reaches. Determine the ti
Angelina_Jolie [31]

Recall that

{v_y}^2-{v_{0y}}^2=2a_y(y-y_0)

At its maximum height y_{\mathrm{max}}, the toy will have 0 vertical velocity, so that

-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}

\implies y_{\mathrm{max}}=1.0\,\mathrm m

For the toy to reach this maximum height, it takes time t such that

\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s

which means it takes twice this time, i.e. t=0.92\,\mathrm s, for the toy to reach its original position.

The velocity of the toy when it falls 1.0 m below its starting point is

{v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)

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\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}

where we took the negative square root because we expect the toy to be moving in the downward direction.

6 0
3 years ago
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