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rjkz [21]
3 years ago
8

Which is an example of the deregulation of a government-regulated natural monopoly? A new law allows consumers to choose between

electricity providers. A new law funds an off-shore wind-power initiative. A new law regulates the number of service interruptions a utility can have. A new law requires underground wires in all high-wind areas.
Physics
1 answer:
expeople1 [14]3 years ago
7 0
<span>An example of the deregulation of a government regulated natural monopoly is where the new ;aw allows consumers to be able to choose between the electricity providers which is the first choice because a deregulation of a government regulated natural monopoly is a way of the rules of having to be remove or reduced when tackling or making use of the government regulated natural monopoly.</span>
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A ferry approaches shore, moving north with a speed of 6.2 m/s relative to the dock. A person on the ferry walks from one side o
jasenka [17]

Speed of Ferry is towards North with magnitude 6.2 m/s

Here if we assume that North direction is along Y axis and East is along X axis then we can say

\vec v_f = 6.2 \hat j

Now a person walk on ferry with speed 1.5 m/s towards east with respect to Ferry

so it is given as

\vec v_{pf} = 1.5 \hat i

also by the concept of relative motion we know that

\vec v_{pf} = \vec v_p - \vec v_f

now plug in all values in it

1.5 \hat i = \vec v_p - 6.2 \hat j

\vec v_p = 1.5 \hat i + 6.2 \hat j

now if we need to find the speed of the person then we need to find its magnitude

so it is given as

v = \sqrt{1.5^2 + 6.2^2}

v = 6.37 m/s

7 0
3 years ago
A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ
ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

\Delta U = q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

For a proton,

q=+e=1.6\cdot 10^{-19}C

And since \Delta V=100 V

The energy gained by the proton is

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

For an alpha particle,

q=+2e=3.2\cdot 10^{-19}C

Therefore, the energy gained is

\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

q=+e=1.6\cdot 10^{-19}C

So the energy gained is the same as the proton:

\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J

6 0
3 years ago
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ryzh [129]
Temperature that will be my answer number 1
7 0
3 years ago
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vazorg [7]
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Help me plz.<br>Show workings​
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