First is melts then it expands next it gets cooler Finally it gains ener. Hope this helps you out.
The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.
<u>Explanation:</u>
According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.
Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.
ΔU = Q+W
Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.
Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.
As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.
No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels
Answer:
x = 17.88[m]
Explanation:
We can find the components of the initial velocity:
![(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7Bo%7D%20%20%3D%2013.3%2Acos%2841.5%29%3D9.96%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%20%3D%2013.3%2Asin%2841.5%29%3D8.81%5Bm%2Fs%5D)
We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.
g = - 9.81[m/s^2]
Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.
![y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2B%28v_%7By%7D%20%29_%7Bo%7D%20%2At-0.5%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5C0%3D1.9%2B%288.81%2At%29-%284.905%2At%5E%7B2%7D%29%5C%5C-1.9%3D8.81%2At%2A%281-0.5567%2At%29%5C%5Ct%3D0%5C%5Ct%3D1.796%5Bs%5D)
With this time we can calculate the horizontal distance:
![x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bx%7D%29_%7Bo%7D%20%2At%5C%5Cx%3D9.96%2A1.796%5C%5Cx%3D17.88%5Bm%5D)