Answer:
1. 610,000 lb ft
2. 490 J
Explanation:
1. First, convert mi/hr to ft/s:
100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s
Now find the kinetic energy:
KE = ½ mv²
KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²
KE = 610,000 lb ft
2. KE = ½ mv²
KE = ½ (5 kg) (14 m/s)²
KE = 490 J
C) A crack in earth's crust where movement occurs. An example of this is the San Andreas Fault.
A) is repulsion
B) continental crust (lighter crust)
D) Hotspot
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
P=IV, where P is power, I is resistance, and V is voltage. Plug in and solve:
P=400(20)
P=8000W
Hope this helps!!
Answer:
wavelength = 0.968 m
frequency = 39.02 Hz
Explanation:
given data
mass = 0.0127 kg
force = 9.33 N
length = 1.93 m
to find out
wavelength and Frequency
solution
we know here linear density that is
linear density =
.........1
linear density =
linear density = 6.5803 ×
kg/m
so
wavelength will be here
wavelength =
..............2
here n = 4 for forth harmonic
wavelength = 
wavelength = 0.968 m
and
frequency will be for 4th normal mode of vibration is
frequency =
..........3
frequency = 
frequency = 1.036269 × 37.654594
frequency = 39.02 Hz