Answer: thermal conductivity
Answer: 80J
Explanation:
According to the first principle of thermodynamics:
<em>"Energy is not created, nor destroyed, but it is conserved." </em>
Then this priciple (also called Law) relates the work and the transferred heat exchanged in a system through the internal energy
, which is neither created nor destroyed, it is only transformed. So, in this especific case of the compressed gas:
(1)
Where:
is the variation in the internal (thermal) energy of the system (the value we want to find)
is the heat transferred out of the gas (that is why it is negative)
is the work is done on the gas (as the gas is compressed, the work done on the gas must be considered positive )
On the other hand, the work done on the gas is given by:
(2)
Where:
is the constant pressure of the gas
is the variation in volume of the gas
In this case the initial volume is
and the final volume is
.
This means:
(3)
Substituting (3) in (2):
(4)
(5)
Substituting (5) in (1):
(6)
Finally:
This is the change in thermal energy in the compression process.
Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.
Now here we will do the components of the weight of the person
given that weight of the person = 500 N
now its components are


now here as we can say that one of the component is balanced here by the normal force perpendicular to plane
while the other component of the weight is balanced by the force applied on the rope
So here the force applied on the rope will be given as


so it apply 300 N force along the inclined plane
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,

Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s