Answer:
The acceleration of the mass is 2 meters per square second.
Explanation:
By Newton's second law, we know that force (
), measured in newtons, is the product of mass (
), measured in kilograms, and net acceleration (
), measured in meters per square second. That is:
(1)
The initial force applied in the mass is:
In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to 
of the initial force. The acceleration of the mass is:
The acceleration of the mass is 2 meters per square second.
Answer:
B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.
Explanation:
B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.
To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.
So things our data are given by
PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by
So the voltage for transformer 2 would be given by,
PART B) To express the number value we proceed to replace with the previously given values, that is to say
Answer:
∆T = Mv^2Y/2Cp
Explanation:
Formula for Kinetic energy of the vessel = 1/2mv^2
Increase in internal energy Δu = nCVΔT
where n is the number of moles of the gas in vessel.
When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas
We say
1/2mv^2 = ∆u
1/2mv^2 = nCv∆T
Since n = m/M
1/2mv^2 = mCv∆T/M
Making ∆T subject of the formula we have
∆T = Mv^2/2Cv
Multiple the RHS by Cp/Cp
∆T = Mv^2/2Cv *Cp/Cp
Since Y = Cp/CV
∆T = Mv^2Y/2Cp k
Since CV = R/Y - 1
We could also have
∆T = Mv^2(Y - 1)/2R k
