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your question here at brainly. Feel free to ask more questions.
<span><span>The
best and most correct answer among the choices provided by the question is </span>B.-2.71 V.</span>
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V
Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V
Since Cu is acting as the anode, the equation needs to be
reversed.
Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V
Ecell= -2.37 V+ (- 0.34 V) = -2.71 V
<span><span>
</span><span>Hope my answer would be a great help for you. </span> </span>
<span> </span>
14 m/s or 50km/h. See the details in the attached picture.
<span>A. Rocket A will travel farther horizontally than rocket B.
This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.</span>
The lemon trees will always be the same so they are the constants. The independent variable is the water amount because it is changing. The dependent variable is the amount of lemons because it is the variable being tested. (If we change the amount of water how many lemons will be produced?)
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
