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3 years ago

Which type of energy resource can be replaced in a short amount of time? 15 Points

Physics

1 answer:

denpristay [2]3 years ago

8 0

Answer:

renewable

Explanation:

Renewable resources are natural resources that can be replaced in a relatively short period of time or are virtually limitless in supply. Renewable energy resources include wind, sunlight, moving water, biomass, and geothermal energy.

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A "590-W" electric heater is designed to operate from 120-V lines.

lukranit [14]

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

I = 4.92 A

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

R = 24.4 Ω

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

P = 495.9 W

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

7 0

3 years ago

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

3 years ago

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the

Novay_Z [31]

Answer:

It will be cut in half

Explanation:

The diffraction of a slit is given by the formula

a sin θ = m where

a = width of the slit,

λ = wavelength and

m = integer that determines the order of diffraction.

Next we divide both sides by a, we have

sin θ = m λ / a

Also, recall that

a’ = 2 a

Then we substitute in the previous equation

2asin θ' = m λ, if divide by 2a, we have

sin θ' = (m λ / 2a).

Now again, from the first equation, we said that sin θ = m λ / a, so we substitute

sin θ ’= sin θ / 2

Then we use trigonometry to find the width, we say

tan θ = y / L

Since the angle is small, we then have

tan θ = sin θ / cos θ

tan θ = sin θ, this then means that

sin θ = y / L

we will then substitute

y’ / L = y/L 1/2

y' = y / 2

this means that when the slit width is doubled the pattern width will then be halved

4 0

3 years ago

An electric current will always follow

MrRa [10]

Answer:

A. The path of least resistance.

Explanation:

ur welcome again ;)

3 0

3 years ago

What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a

astra-53 [7]

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section area $A = (2.00\times10^{-3})^2\ m$

(I). We need to calculate the area of copper wire

Using formula of area

$A=\pi r^2$

$A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}$

$R=0.0742\ \Omega$

(II). We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}$

$R=1.75\ \Omega$

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

8 0

3 years ago