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 
.)
.
The First Law describes how an object acts when no force is acting upon it. So, rockets stay still until a force is applied to move them. Likewise, once they're in motion, they won't stop until a force is applied. Newton's Second Law tells us that the more mass an object has, the more force is needed to move it. A larger rocket will need stronger forces (eg. more fuel) to make it accelerate. The space shuttles required seven pounds of fuel for every pound of payload they carry. Newton's Third Law states that "every action has an equal and opposite reaction". In a rocket, burning fuel creates a push on the front of the rocket pushing it forward.
Answer:
68 db
Explanation:
Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.
Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,
β(db)=10×㏒(2)
=3 db
So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
