Answer:
Impulse, |J| = 0.6716 kg-m/s
Force, F = 63.35 N
Explanation:
It is given that,
Mass of the baseball, m = 0.146 kg
Initial speed of the ball, u = 15.3 m/s
Final speed of the ball, v = 10.7 m/s
To find,
(a) The magnitude of this impulse.
(b) The magnitude of the average force of the glass on the ball.
Solution,
(a) Impulse of an object is equal to the change in its momentum. It is given by :
J = -0.6716 kg-m/s
or
|J| = 0.6716 kg-m/s
(b) Another definition of impulse is given by the product of force and time of contact.
t = 0.0106 s
F = 63.35 N
Hence, this is the required solution.
Answer:
the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg
Explanation:
To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables
Mathematically this can be determined as
Where
Temperature at inlet of turbine
Temperature at exit of turbine
Pressure at exit of turbine
Pressure at exit of turbine
The steady flow Energy equation for an open system is given as follows:
Where,
m = mass
m(i) = mass at inlet
m(o)= Mass at outlet
h(i)= Enthalpy at inlet
h(o)= Enthalpy at outlet
W = Work done
Q = Heat transferred
v(i) = Velocity at inlet
v(o)= Velocity at outlet
Z(i)= Height at inlet
Z(o)= Height at outlet
For the insulated system with neglecting kinetic and potential energy effects
Using the relation T-P we can find the final temperature:
From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK
the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg
B. Orbit. The planets orbit the sun, the moon orbits earth, etc.
Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s