Answer:
b) 7.00
Explanation:
N( t ) = -20( t - 5 )²
dN/ dt = -20 x 2 ( t - 5 )
For maximum N ( depth )
dN/dt = 0
- 40 ( t - 5 ) = 0
t = 5
So at 2 + 5 = 7 .00 am depth of water reaches its maximum.
Answer:
density of cube =11.605 g/cm³
Explanation:
density of a substance is the mass per unit volume of that substance.
the density of a substance = 
volume of a cube = l³,
l = 19.0mm , lets convert mm to cm
1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm
length of cube =1.9cm
volume of cube = 1.9³
density of cube = 
density of cube =11.605 g/cm³
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:
Flip the cell.
Explanation:
This reverses direction of energy transfer.
Alternatively, flip ammeters and voltmeters to give negative readings.
what do penguins eat for lunch?
Ice-burgers!
Answer:
Work done by a tug boat, W = 1.735 x 10⁸ J
Explanation:
Given,
The of each tugboat, F = 1.5 x 10⁶ N
The angle of each tugboat forms with the resultant force, θ = 19°
The displacement of the supertanker, s = 710 m
The individual tugboat will be responsible for the displacement, d = 710/2
= 355 m
The displacement component in each tugboat direction = 355 · sin θ meter
Therefore, the work done by each tugboat is
W = F x S joules
Substituting the values in the above equation
W = 1.5 x 10⁶ x 355 · sin θ
= 1.735 x 10⁸ J
Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J