Answer:
39 m/s
Explanation:
Hi!
The equation of motion of the ball is:

Where y_0 is the initial horizontal position, which we will take as zero.
v_0y is the initial velocity and t is the time
To find the initial velocity we will set y(7.96s) = 0

Answer:
The acceleration of the box is 3 m/s²
Explanation:
Given;
mass of the box, m = 12 kg
horizontal force pulling the box forward, Fx = 48 N
frictional force acting against the box in opposite direction, Fk = 12 N
The net horizontal force on the box, F = 48 N - 12 N
The net horizontal force on the box, F = 36 N
Apply Newton's second law of motion to determine the acceleration of the box;
F = ma
where;
F is the net horizontal force on the box
a is the acceleration of the box
a = F / m
a = 36 / 12
a = 3 m/s²
Therefore, the acceleration of the box is 3 m/s²
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


T = tension force in the rope in upward direction
m = mass of the box attached at end of rope = 56 kg
W = weight of the box in downward direction due to gravity
a = acceleration of the box in upward direction = 5.10 m/s²
weight of the box is given as
W = mg
inserting the values
W = (56) (9.8)
W = 548.8 N
force equation for the motion of the box is given as
T - W = ma
inserting the values
T - 548.8 = (56) (5.10)
T = 834.4 N
C liquid at room temperature