Answer: Rc = 400 Ω and Rb = 57.2 kΩ
Explanation:
Given that;
VCE = 5V
VCC = 15 V
iC = 25 mA
β = 100
VD₀ = 0.7 V
taking a look at the image; at loop 1
-VCC + (i × Rc) + VCE = 0
we substitute
-15 + ( 25 × Rc) + 5 = 0
25Rc = 10
Rc = 10 / 25
Rc = 0.4 k
Rc = 0.4 × 1000
Rc = 400 Ω
iC = βib
25mA = 100(ib)
ib = 25 mA / 100
ib = 0.25 mA
ib = 0.25 × 1000
ib = 250 μAmp
Now at Loop 2
-Vcc + (ib×Rb) + VD₀ = 0
-15 (250 × Rb) + 0.7 = 0
250Rb = 15 - 0.7
250Rb = 14.3
Rb = 14.3 / 250
Rb = 0.0572 μ
Rb = 0.0572 × 1000
Rb = 57.2 kΩ
Therefore Rc = 400 Ω and Rb = 57.2 kΩ
Answer:
, 
Explanation:
Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to
. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:
Principle of Mass Conservation

First Law of Thermodynamics

This 2 x 2 System can be reduced into one equation as follows:

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:
Inflow (Superheated Steam)

The mass flow rate can be calculated by using this expression:


Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:


The enthalpy rate at outflow is:


Answer: all you need to to is go to help me with this question.cm
Explanation:
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The solution has been attached to the portal
Answer:
Area under the strain-stress curve up to fracture gives the toughness of the material.
Explanation:
When a material is loaded by external forces stresses are developed in the material which produce strains in the material.
The amount of strain that a given stress produces depends upon the Modulus of Elasticity of the material.
Toughness of a material is defined as the energy absorbed by the material when it is loaded until fracture. Hence a more tough material absorbs more energy until fracture and thus is excellent choice in machine parts that are loaded by large loads such as springs of trains, suspension of cars.
The toughness of a material is quantitatively obtained by finding the area under it's stress-strain curve until fracture.