Answer:
v = 7121.3 m/s
Explanation:
As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed
so here we will have
here we know that
r = orbital radius = 6370 km + 1482 km
also we know that
now we will have
Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T
in the cycle is twice the minimum absolute temperature T
in the cycle
T = 0.5T
now, we find the efficiency of the Carnot cycle engine
η
= 1 - T/T
η = 1 - T/0.5T
η = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η = 1 - W
/Q
where W is net work done, Q is is the heat supplied
we substitute
0.5 = 60 / Q
Q = 60 / 0.5
Q = 120 kJ
Now, we apply the first law of thermodynamics to the system
W = Q - Q
60 = 120 - Q
Q = 60 kJ
now, the amount of heat rejection per kg of steam is;
q = Q/m
we substitute
q = 60/0.025
q = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q = h
= 2400 kJ/kg
now, at h = 2400 kJ/kg from saturated water tables;
T = 40 + ( 45 - 40 ) ( 
)
T = 40 + (5) × (0.5)
T = 40 + 2.5
T = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C
The
sun is a ball of hot gases containing different kinds of elements at different
cores. It has a very high temperature that radiates all throughout the Milky
Way galaxy. The sun has three main parts; photosphere, chromospheres
and corona. The outer core of a star located at the chromospheres contains
mostly of hydrogen. Inside the hydrogen is helium then carbon, oxygen, neon,
magnesium silicon and the inert gas. The photosphere is scattered by the loose electrons in the corona’s plasma.
