Complete Question
A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.
(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).
J
(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)
MW
Answer:
The heat transferred is 
The power is 
Explanation:
From the question we are told that
Mass of the water per second is 
The initial temperature of the water is 
The boiling point of water is 
The final temperature 
The latent heat of vapourization of water is 
The specific heat of water 
The specific heat of stem is 
Generally the heat needed each second is mathematically represented as
![Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)]](https://tex.z-dn.net/?f=Q%20%3D%20m%5Bc_w%20%28T_i%20-%20T_b%29%20%2B%20m%2A%20c__%7BL%7D%7D%20%20%2B%20m%2A%20c__%7BS%7D%7D%20%28T_f%20-%20T_b%29%5D)
Then substituting the value
![Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]]](https://tex.z-dn.net/?f=Q%20%3D%20m%5Bc_w%20%5BT_i%20-%20T_b%5D%20%2B%20c__%7BL%7D%7D%20%20%2B%20C__%7BS%7D%7D%20%5BT_f%20-%20T_b%5D%5D)
![Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]]](https://tex.z-dn.net/?f=Q%20%3D%201917%20%5B%284184%29%20%5B100%20-%2035%5D%20%2B%20%5B2256%20%2A%2010%5E3%5D%20%20%2B%5B1520%5D%20%20%5B450%20-%20100%5D%5D)
![Q = 1917 * [3.05996 * 10^6]](https://tex.z-dn.net/?f=Q%20%3D%201917%20%2A%20%5B3.05996%20%2A%2010%5E6%5D)

The power required is mathematically represented as

From the question 
So



-- What's the volume of a cylinder with radius=1m and height=55m ?
( Volume of a cylinder = π R² h )
-- How much does that volume of water weigh ?
1 liter of water = 1 kilogram of mass
Weight = (mass) x (acceleration of gravity)
-- What's the area of the bottom of that 1m-radius cylinder ?
Pressure = (force) / (area)
Answer:
A. 
B. 
C. 
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

is the capacitance,
is the common plate area,
is the plate separation and
is the permittivity of the material between the plates.
For air or free space,
is
called the permittivity of free space. In general,
where
is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum,
.
The energy stored in a capacitor is the average of the product of its charge and voltage.

Its charge,
, is related to its capacitance by
(this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for
,

A. Substituting for
in
,

B. When the distance is
,


C. When the distance is restored but with a dielectric material of dielectric constant,
, inserted, we have

Answer:
v_{f} = 115.95 m / s
Explanation:
This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions
Thrust =
-v₀ = v_{e} 
where v_{e} is the velocity of the gases relative to the rocket
let's apply these expressions to our case
the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units
M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg
The final mass is the mass of the engines + the mass of the rocket
M_{f} = 25.5 +54.5 = 80 g = 0.080 kg
thrust and duration of ignition are given
thrust = 5.26 N
t = 1.90 s
Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear
thrust = v_{e}
v_{e} = thrust
v_{e} = 5.26
v_{e} = - 786.93 m / s
the negative sign indicates that the direction of the gases is opposite to the direction of the rocket
now we look for the final speed of the rocket, which as part of rest its initial speed is zero
v_{f}-0 = v_{e} 
we calculate
v_{f} = 786.93 ln (0.0927 / 0.080)
v_{f} = 115.95 m / s
A discovery not the others