Answer:
Explanation:
The process is isothermic,  as P V = constant . 
work done = 2.303 n RT log P₁ / P₂
= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ 
= 300.5k J
This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature. 
heat supplied  = 300.5k J
specific volume is volume per unit mass 
v / m 
pv = n RT 
pv  = m / M  RT
v / m = RT / p M 
specific volume = RT / p M 
option B is correct. 
 
        
             
        
        
        
The answer in Meters is going to to 1265.341
        
                    
             
        
        
        
Answer:
40 N
Explanation:
We are given that 
Speed of system is constant 
Therefore, acceleration=a=0
Tension applied on block B=T=50 N
Friction force=f=10 N
We have to find the friction force acting on block A.
Let T' be the tension in string connecting block A and block B and friction force on block A be f'.
For Block B

Where  =Mass of block B
=Mass of block B
Substitute the values 


For block A

Where  Mass of block A
Mass of block A
Substitute the values 


Hence, the friction force acting on block A=40 N
 
        
             
        
        
        
Answer:
Gallium
Explanation:
Gallium is one such element used as a do/pant in a p-type semiconductor. 
A do/pant is an impurity added to a semi-conductor used to alter its properties. Semi-conductors have a wide range of applications. They will conduct heat and electricity only under certain conditions. This property is highly desirable and find a wide application in electronics. 
For p-type conductors, they are best do/ped with elements with 3 valence electrons. These are group 3 elements. From the choices, only gallium belongs to this group. 
Other elements given are good do/pants for n-type semiconductors. They have 5 valence electrons. 
 
        
             
        
        
        
Answer:
a = 2 [m/s²]
Explanation:
To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.
We can use the following equation.

where:
Vf = final velocity = 11 [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration [m/s²]
t = time = 3 [s]
![11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]](https://tex.z-dn.net/?f=11%20%3D%205%20%2B%20a%2A3%5C%5C6%3D3%2Aa%5C%5Ca%3D%202%5Bm%2Fs%5E%7B2%7D%20%5D)