Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.
Answer:
a. 12 m/s² down
Explanation:
Acceleration has units of length per time squared. Acceleration is a vector, so it also has a direction.
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m
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
Answer:
1 kg
Explanation:
The container has negligible mass and no heat is loss to the surrounding.
Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg
Using the formula:
Quantity of heat gain by ice = Quantity of heat loss by water
Quantity of heat gain by ice = mass of ice × heat of fusion of ice + mass of water × specific heat capacity of water = (0.4 × 3.36 × 10^ 5) + (0.4 × 4200 × (26- (-29) = 13.44 × 10^4 + 9.24 × 10^ 4 = 22.68 × 10^4 J
Quantity of heat loss by water = m2cΔT
Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)
since heat gain = heat loss
22.68 × 10^4 = 226800 m2
divide both side by 226800
226800 / 226800 = m2
m2 = 1 kg