Problem definition

MakcuM [25]

In surface mining, stripping ratio or strip ratio refers to the amount of waste (or overburden) that must be removed to release a given ore quantity.

8 0

2 years ago

Read 2 more answers

A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

PIT_PIT [208]

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0

3 years ago

A 3.7 g mass is released from rest at C which has a height of 1.1 m above the base of a loop-the-loop and a radius of 0.2 m . Th

Dmitrij [34]

Answer:

Normal force = 0.326N

Explanation:

Given that:

mass released from rest at C = 3.7 g = 3.7 × 10⁻³ kg

height of the mass = 1.1 m

radius = 0.2 m

acceleration due to gravity = 9.8 m/s²

We are to determine the normal force pressing on the track at A.

To to that;

Let consider the conservation of energy relation; which says:

mgh = mgr + 1/2 mv²

gh = gr + 1/2 v²

gh - gr = 1/2v²

g(h-r) = 1/2v²

v² = 2g(h-r)

However; the normal force will result to a centripetal force; as such, using the relation

N =mv²/r

replacing the value for v² = 2g(h-r) in the above relation; we have:

Normal force = 2mg(h-r)/r

Normal force = 2 × 3.7 × 10⁻³ × 9.8 ( 1.1 - 0.2 )/ 0.2

Normal force = 0.065268/0.2

Normal force = 0.32634 N

Normal force = 0.326N

6 0

3 years ago

You are designing a 200mm long x 100mm wide x 50mm deep rectangular housing, with a wall width of 1.5 mm, and 1 degree draft. Yo

Nimfa-mama [501]

365gpa I don’t even know what it is but this is with 10 point and I need it

6 0

2 years ago

A product whose total work content time = 50 min is assembled on a manual production line at a rate of 24 units per hour. From p

valentinak56 [21]

Answer:

a)Cycle time = 2.37 min

b)Numbers of workers =21

c)Stations on the line =24

Explanation:

Given that

Total work content time(TWC) = 50 min

Production rate Rp= 24 units/hr

manning level will be close =1.5

Line balancing efficiency =0.94

a)

Cycle time

$T_c=\dfrac{60E}{R_P}$

$T_c=\dfrac{60\times 0.95}{24}$

Cycle time = 2.37 min

b)

Numbers of workers ,W

$W=\dfrac{TWC}{T_c}$

$W=\dfrac{50}{2.37}$

W= 21

Numbers of workers =21

c)

Stations on the line(n)

Lets find service time Ts

Ts = Cycle time - Time for repositioning

Ts = Tc- Tr

Ts= 2.37 - 9/ 60 min

Ts= 2.22 min

We know that efficiency

$\eta=\dfrac{TWC}{n.T_s}$

$0.94=\dfrac{50}{n\times 2.22}$

n=23.94 ⇒n=24

n=24

Stations on the line =24

7 0

3 years ago