Answer:
45.3 MN
Explanation:
The forging force at the end of the stroke is given by
F = Y.π.r².[1 + (2μr/3h)]
The final height, h is given as h = 100/2
h = 50 mm
Next, we find the final radius by applying the volume constancy law
volumes before deformation = volumes after deformation
π * 75² * 2 * 100 = π * r² * 2 * 50
75² * 2 = r²
r² = 11250
r = √11250
r = 106 mm
E = In(100/50)
E = 0.69
From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula
F = Y.π.r².[1 + (2μr/3h)]
F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]
F = 35.3 * [1 + 0.2826]
F = 35.3 * 1.2826
F = 45.3 MN
Answer:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
Flip flops are not required
Answer:
option c is correct
47.2%
Explanation:
given data
consisting of refrigerant = 134 a
volume V = 0.01 m³/kg
pressure P = 1MPa = 1000 kPa
to find out
quality of the R 134a
solution
we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a
that is
Vf = 0.0008701 m³/kg
Vv = 0.0203 m³/kg
so we will apply here formula that is
quality = (V - Vf) / (Vv - Vf) ............1
put here value
quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )
quality = 0.4698
so quality is 47 %
SO OPTION C IS CORRECT
Answer:

Explanation:
<u>Projectile Motion</u>
In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

Where
is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

The horizontal and vertical distances are, respectively:


The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

Let's solve for 

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:


Or equivalently:

Given Vo=37 m/s and R=70 m


And
