4. Wheel grinders need be equipped with a(n):

Select four items that an industrial engineer must obtain in order to practice in the field.

alex41 [277]

Answer:

Professional engineering license

Bachelor's degree

Computer science classes

job recommendations

3 0

2 years ago

Read 2 more answers

La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación

gregori [183]

Answer:

I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

I₀ = $\frac{I}{sin \ \theta }$

I₀ = 1600 /sin 53

I₀ = 2003.42 lx

for when the angle is θ = 37º

I = 2003.42 sin 37

I = 1205.69 Lx

6 0

2 years ago

When would working with machinery be a common type of caught-in and caught-between hazard?

tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

6 0

3 years ago

A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.

Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = $\frac{PL}{AE}$ ................1

here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = $\frac{PL}{AE}$

δ = $\frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}$

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0

3 years ago

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg

b) 504 kJ

c) 514 kJ

Explanation:

Given

-The mass of C_o2 = 1 kg

-The volume of the tank V_tank = 1 m^3

-The added energy E = 14 W

-The time of adding energy t = 10 s

-The increase in specific internal energy Δu = +10 kJ/kg

-The change in kinetic energy ΔKE = 0 and The change in potential energy

ΔPE =0

Required

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.

(c)The energy transferred by the heat transfer W in kJ and the direction of

the heat transfer.

Assumption

-Quasi-equilibrium process.

Solution

(a) The volume and the mass doesn't change then, the specific volume is constant.

v= V_tank/m ---> 1/1= 1 m^3/Kg

(b) The added work is defined by.

W =E * t ---> 14 x 10 x 3600 x 10^-3 = 504 kJ

(c) From the first law of thermodynamics.

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0

3 years ago