Hi there!
We can use the following (derived) equation to solve for the final velocity given height:
vf = √2gh
We can rearrange to solve for height:
vf² = 2gh
vf²/2g = h
Plug in the given values (g = 9.81 m/s²)
(13)²/2(9.81) = 8.614 m
We can calculate time using the equation:
vf = vi + at, where:
vi = initial velocity (since dropped from rest, = 0 m/s)
a = acceleration (in this instance, due to gravity)
Plug in values:
13 = at
13/a = t
13/9.81 = 1.325 sec
Answer:
Approximately
.
Explanation:
Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (
) is equal to
.
There are two half-reactions in this question.
and
. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of
should be positive.
In this case,
is positive only if
is the reaction takes place at the cathode. The net reaction would be
.
Its cell potential would be equal to
.
The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:
,
where
is the number moles of electrons transferred for each mole of the reaction. In this case the value of
is
as in the half-reactions.
is Faraday's Constant (approximately
.)
.
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:

ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Finally, you obtain for E:

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
That ratio is called"efficiency". It doesn't need to be a percent.
It can just as well be a fraction or a decimal number.
First,

where
is density,
is mass, and
is volume. We can compute the volume of the roll:


When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness
. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have

where
is the given area, so


If we're taking significant digits into account, the volume we found would have been
, in turn making the thickness
.