Each electron has a charge of
In this problem, the total charge is
Therefore, the number of electrons contained in this total charge will be given by the total charge divided by the charge of a single electron:
<span>In order best to find out whether the obtained experimental results are worth mor etime and resources the group of scientists should present their results (could be done also in poster session) to other scientists in order to hear their opinion and get a feedback. The shoudl also ask another researchers to redo the experiments and to compare the results. </span>
Answer:
The value is 
Explanation:
From the question we are told that
The molar mass of hydrazine is 
The initial temperature is 
The final temperature is 
The specific heat capacity is ![c_h = 0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)](https://tex.z-dn.net/?f=c_h%20%20%3D%20%200.099%20%5BkJ%2F%28mol%20K%29%5D%20%3D%200.099%20%2A10%5E3%20J%2F%28mol%2FK%29)
The power available is 
The mass of the fuel is 
Generally the number of moles of hydrazine present is
=> 
=> 
Generally the quantity of heat energy needed is mathematically represented as
=> 
=> 
Generally the time taken is mathematically represented as
=> 
=> t = 2480505.6377 s
Converting to hours
=>
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
