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1 year ago

What name is given to the system that distributes electricity from power stations around the UK?

Physics

1 answer:

steposvetlana [31]1 year ago

5 0

Answer:

The National Grid is the system operator of Great Britain's electricity and gas supply. This includes England, Scotland and Wales. It is the company that manages the network and distribution of electricity and gas that powers all our homes and businesses.

Explanation:

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Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this

Crank

Energy of gamma rays is given by equation

$E = h\nu$

here we know that

h = Planck's constant

$\nu = frequency$

now energy is given as

$E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}$

$E = 7.52 \times 10^{-13} J$

now by above equation

$E = h\nu$

$7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu$

$\nu = 1.14 \times 10^{21} Hz$

now for wavelength we can say

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}$

$\lambda = 2.63 \times 10^{-13} m$

3 0

2 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

2 years ago

Explain the difference between evaporation and boiling

azamat

Well evaporation is a gradual vaporization of a liquid on the surface whereas boiling is a rapid vaporization of a liquid only when it is heated to its boiling point.

7 0

3 years ago

Read 2 more answers

Velocity question...is it A B Cor D

Ket [755]

In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

hope this helps :D

6 0

2 years ago

Read 2 more answers

If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surfac

Andru [333]

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

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2 years ago