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3 years ago

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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

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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The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

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The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

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The wavelength range for the corresponding frequency.

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The speed of sound is given by the following relation as :

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$\lambda_2=\dfrac{v}{f_2}$

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A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

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Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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