Heat required to raise the temperature of a given system is
here we know that
m = mass
s = specific heat capacity
= change in temperature
now as we know that
mass of wood = 5 kg
mass of aluminium pan = 2 kg
change in temperature = 45 - 20 = 25 degree C
specific heat capacity of wood = 1700 J/kg C
specific heat capacity of aluminium = 900 J/kg C
now here we will find the total heat to raise the temperature of both
So heat required to raise the temperature of the system is 257500 J
<span><span>anonymous </span> 4 years ago</span>Any time you are mixing distance and acceleration a good equation to use is <span>ΔY=<span>V<span>iy</span></span>t+1/2a<span>t2</span></span> I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890.
For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time.
Add the two times together for the total.
The alternative is to calculate the initial and final velocity so that you have more information to work with.
During upward projection the final velocity is zero, and the gravitational acceleration is -10 m/s² (against the gravity).
Therefore; using the equation;
S = 1/2gt² + ut
Where s is the height h, g is gravitational acceleration, and t is the time and u is the initial velocity u, is 16 ft/s.
Thus; h= 1/2(-10)t² + 16t
We get; h = -5t² + 16t
Therefore; the quadratic equation is 5t² - 16t + h =0
The force is opposite to the displacement
To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,
Here,
k = Coulomb's constant
= Charge at each object
r = Distance between them
As the distance is doubled so,
Therefore the factor is 1/4
