Answer:
a ) 
b) 
Explanation:
given data:
pressure ration rp = 12
inlet temperature = 300 K
TURBINE inlet temperature = 1000 K
AT the end of isentropic process (compression) temperature is



AT the end of isentropic process (expansion) temperature is



isentropic work is given as

w = 1.005(610.18 - 300)
w = 311.73 kJ/kg
w(turbine) = 1.005( 1000 - 491.66)
w(turbine) = 510.88 kJ/kg
a) mass flow rate for isentropic process is given as


b) actual mass flow rate uis given as


v = initial velocity of launch of the stone = 12 m/s
θ = angle of the velocity from the horizontal = 30
Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.
v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s
a = acceleration of the stone = - 9.8 m/s²
t = time of travel = 4.8 s
Y = vertical displacement of stone = vertical height of the cliff = ?
using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²
Y = - 84.1 m
hence the height of the cliff comes out to be 84.1 m
Answer:
B - Earth's path around the Sun
Explanation:
Answer:
Shawn's speed relative to Susan's speed = 10 mph
Resultant velocity = 82.32 mph
Explanation:
The given data :-
i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.
ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.
iii) The speed of both Susan and Shawn is relative to earth.
iv) The angle between Susan in north and Shawn in east is 90°.
We have to find Shawn's speed relative to Susan's speed.
v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph
Resultant velocity,

v = 82.32 mph
It will be cloudy and there will be rain.
Hope this helps