Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow

The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by

The density of the flow at the exit is 2.2721 kg/m³
Answer: option d: The nucleus of Atom Q is more stable than the nucleus of Atom P.
Explanation:
Atom P is radioactive and disintegrates, it emits beta particles (high speed electrons or positrons) because it is not stable. On disintegration, it forms a stable Atom Q which is non-radioactive and thus it does not disintegrates further.
Thus, the correct option is only d. The nucleus of Atom Q is more stable than the nucleus of Atom P.
(a)
The work done on the projectile is 9375 joule.
The work on the projectile is calculated as
W=F×d
=1250×7.5
=9375 joule
(b)
The speed of the projectile after 7.5 m is 27.38 m/s
First we need to find out the acceleration of the projectile
F=m×a
1250=25×a
a=50 m/
Now the velocity of the projectile after 7.5 m is calculated as
v^2=u^2+2a×s
v^2=0+2×50*7.5
v=27.38 m/s