The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Answers
Answer:
Magnitude of force P = 25715.1517 N
Explanation:
Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.
To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.
Proof -
Given that,
Diameter = 12 mm = 0.012 m
Length = 0.6 m
[tex]\theta[/tex] = 0.015°
Youngs modulus of elasticity of 34 stainless steel is 193 GPa
Now,
By applying the conditions of equilibrium, we have
∑fₓ = 0, ∑[tex]f_{y}[/tex] = 0, ∑M = 0
If ∑[tex]M_{A}[/tex] = 0
⇒[tex]F_{BC}[/tex]×0.9 - P × 0.6 = 0
⇒[tex]F_{BC}[/tex]×3 - P × 2 = 0
⇒[tex]F_{BC}[/tex] = [tex]\frac{2P}{3}[/tex]
If ∑[tex]M_{B}[/tex] = 0
⇒[tex]F_{AD}[/tex]×0.9 = P × 0.3
⇒[tex]F_{AD}[/tex] ×3 = P
⇒[tex]F_{AD}[/tex] = [tex]\frac{P}{3}[/tex]
Now,
Area, A = [tex]\frac{\pi }{4} X (0.012)^{2}[/tex] = 1.3097 × 10⁻⁴ m²
We know that,
Change in Length , [tex]\delta[/tex] = [tex]\frac{P l}{A E}[/tex]
Now,
[tex]\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 9.1626 × 10⁻⁹ P
[tex]\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }[/tex] = 1.83253 × 10⁻⁸ P
Given that,
[tex]\theta[/tex] = 0.015°
⇒[tex]\theta[/tex] = 2.618 × 10⁻⁴ rad
So,
[tex]\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}[/tex]
⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9
⇒P = 25715.1517 N
∴ we get
Magnitude of force P = 25715.1517 N
The magnitude of Force P is; P = 25715.15 N
What is the magnitude of the force?
If we draw a free body diagram of the rigid beam system, then for beam AB we can take moments in the following manner;
Taking moments about point A, we have;
(F_bc * 0.9) - P(0.6) = 0
F_bc = ²/₃P
Taking moments about B gives;
P(0.3) - F_ad * 0.9 = 0
F_ad = ¹/₃P
Normal stress for BC is;
σ_bc = F_bc/A_bc
σ_bc = (²/₃P)/(π * 0.006²)
σ_bc = (²/₃P)/(1.131 × 10⁻⁴) N/m²
σ_ad = (¹/₃P)/(π * 0.006²)
σ_ad = (¹/₃P)/(1.131 × 10⁻⁴) N/m²
We know that;
Elongation is; ΔL = PL/AE = (P/A) * (L/E)
Where E for 304 stainless steel is 193 GPa = 193 × 10⁹ Pa
Thus;
ΔL_bc = (²/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))
ΔL_bc = 1.83253P × 10⁻⁸
Likewise;
ΔL_ad = (¹/₃P)/(1.131 × 10⁻⁴) * (0.6/(193 × 10⁹))
ΔL_ad = 9.1626P × 10⁻⁹ m
Converting the beam tilt angle from degrees to radians gives;
θ = 0.015° = 0.00026179939 rads
Using small angle analysis, we can say that;
θ = (ΔL_bc - ΔL_ad)/36
θ = P((1.83253P × 10⁻⁸) - (9.1626P × 10⁻⁹))/36
Solving gives P = 25715.15 N
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Related Questions
Derive the next state equations for each type (D, T, SR, and JK) of basic memory element. The next state equation is a symbolic equation describing the next state (Q ) as a function of the inputs (D,T,SR, or JK) and state (Q). In order to determine the next state equations for a a JK memory element, build a 3-variable Kmap with Q, J, and K as the inputs. The entries in the Kmap should be Q . Solving this Kmap will yield the next state equation. Show all work for full credit.
Answers
Answer:
Attached below is the derived next state equations
Explanation:
Attached below is the derived next state equations
used for the solution of the given problem.
4. What element is missing from construction drawings?
A. Physical arrangement of specific electrical equipment
B. Electrical layout
C. Electrical connections
D. Side elevation views
Answers
Answer:
C Electrical Connections
Explanation:
In reading says . However, electrical
connections aren’t shown in construction drawings.
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.
Required:
Determine the normal stress σ in the steel tube.
Answers
Answer:
Explanation:
From the information given:
[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]
The total load is distributed across both the rod and tube:
[tex]P = P_1+P_2 --- (1)[/tex]
Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.
[tex]\delta_1=\delta_2[/tex]
[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]
[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]
[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]
[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]
[tex]P_2 = 6.6212 \ P_1[/tex]
Replace [tex]P_2[/tex] into equation (1)
[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]
Finally, to determine the normal stress in aluminum rod:
[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]
[tex]\sigma_1 = - 23.523 \ MPa}[/tex]
Thus, the normal stress = 23.523 MPa in compression.
An array of eight aluminum alloy long fins, each 3 mm wide, 0.4 mm thick, and 40 mm long, is used to cool a transistor. When the base is at 340 K and the ambient air is at 300 K, how much power do they dissipate if the combined convection and radiation heat transient coefficient is estimated to be 8 W/m2K? The alloy has a conductivity of 175 W/mKand the heat transfer from the tip is negligible.
Answers
Answer:
0.08704 W
Explanation:
converting the mm to m (1000mm = 1m)
cross-sectional area of the fins, Ac = (0.003) (0.0004) = 0.0000012m^2
The wetted perimeter of the cross-section, P = 2 (0.003 + 0.0004) = 0.0068m
Thickness of solid in direction of heat flow, B^2 = (heat transient coefficient, h) (The wetted perimeter of the cross-section, P) ÷ (Thermal conductivity, k) (cross-sectional area of the fins, Ac)
B^2 = (8 W/m2K)(0.0068m) ÷ (175 W/mK)(0.0000012m^2)
=259.0476m^-2
B= square root of the result
B = 16.09m^-1
we now look for:
The Coordinate, x = B, multiplied by Length, L
x = (16.09m^-1) (0.04m) = 0.6436
finding the side area of a fin = P multiplied by Length, L
= 0.0068m X 0.04m = 0.000272m^2
Neglecting inefficiency, assuming the fins are all 100% efficient, the power they would dissipate =
h, Heat-transfer coefficient (PL) (temperature of at the base - temperature at the ambient air)
= (8) (0.000272m^2)(340 K- 300k)
= 0.08704 W
How does distribution add value to goods and services being sold,
including intellectual property?
Answers
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Explanation:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.
Answer:
Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.
Explanation:
hope it helps <33
Poly(cis-1,4-isoprene), or natural rubber (NR), has a tendency crystallize. The Tm of this polymer is slightly below room temperature, although lightly-crosslinked NR can partially crystallize at room temperature when stretched. Apparently, Tm is elevated upon stretching which allows for crystallization above the Tm of the unstretched polymer. Explain.
Answers
Answer:
Explanation:
Crystalline melting temperature (Tm) is termed as the temperature required for a crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).
Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.
For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.
The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.
The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:
1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.
2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.
3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.
Which one of the following answer options are your employers responsibility
Answers
Where are your answer options?
Answer:
Implement a hazard communication program
Explanation: i took the quiz
The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter of the rod be not to deform
Answers
Answer:
r = 1.922 mm
Explanation:
We are given;
Yield stress; σ = 250 MPa = 250 N/mm²
Force; F = 29 KN = 29000 N
Now, formula for yield stress is;
σ = F/A
A = F/σ
Where A is area = πr²
Thus;
r² = 2900/250π
r² = 3.6924
r = √3.6924
r = 1.922 mm
How do you explain the application of regulations in locations containing baths, showers and electric floor heating, including the requirements needed?
Answers
Answer:
The application of regulations in locations are very important.
Explanation:
The application of regulations in locations are very important in order to gain more benefit from it because people choose those places that are well regulated and having more facilities. If the location has baths, showers, electric floor heaters and other necessities so the people prefer the place over another and increase of clients occurs which give more benefits to the place owners.
Where do greywater pipes generally feed into?
-Vent stack
-Water heater
-Waste stack
-Main supply
Answers
Answer:
c Waste stack
Explanation:
A 4 stroke over-square single cylinder engine with an over square ratio of 1.1,the displacement volume of the engine is 245cc .The clearance volume is 27.2cc the bore of this engine is ?
Answers
Answer:
10.007
Explanation:
Assuming we have to find out the compression ratio of the engine
Given information
Cubic capacity of the engine, V = 245 cc
Clearance volume, V_c = 27.2 cc
over square-ratio = 1.1
thus,
D/L = 1.1
where,
D is the bore
L is the stroke
Now,
Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]
plugging values we get
245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]
Solving we get D =7 cm
therefore, L= 7/1.1 =6.36 cm
Now,
the compression ratio is given as:
r =(V+V_c)/V_c
on substituting the values, we get
r = (245+27.2)/27.2 =10.007
Hence, Compression ratio = 10.007
the removed soil at an excavation site is also called spoil?
Answers
Answer:
True, That is correct. Soil removed from an excavation site is indeed called spoil.
Spoil definition: The waste material (such as soil) brought up during the course of an excavation.
Hope I helped! Sorry if not. Have a wonderful week and follow me for more help! Remember your worth and love yourself! Adíos! ;D
Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Thus, The debris that is dug up during an excavation, such as soil. Depending on the location and the type of excavation, the composition of the spoil material can change.
It could consist of dirt, gravel, boulders, debris, or other substances that were removed or moved during the excavation process.
Typically, the spoil is put aside and used for a variety of tasks, including backfilling, grading, or disposal, as necessary for the project and permitted by local laws.
Thus, Yes, that is true. It is true that spoil refers to soil excavated from an excavation site.
Learn more about Soil, refer to the link:
https://brainly.com/question/31227835
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Forces always act in equal and opposite pairs
Answers
A power winch is designed to raise a 4,961 N load 10 meter in 2 minutes. The winch is designed with a 283 mm diameter drum taking up the wire lifting the load. The drum will be connected to a 9 rpm motor through a gearbox. What is the minimum torque (Nm) that the motor shaft coupling should be designed to transmit
Answers
Answer:
T = 438.87 N.m
Explanation:
The power required to raise the 4961 N load in 10 meters for 2 minutes is:
[tex]P = \dfrac{4961*10}{2*60}\\ \\ P = 413.42 Nm/sec[/tex]
P = Torque × W
[tex]413.42 = T \times \dfrac{2 * \pi*9}{60}[/tex]
[tex]413.42 = T \times0.942 \\ \\ T = \dfrac{413.42}{0.942}[/tex]
[tex]T = \dfrac{413.42}{0.942}[/tex]
T = 438.87 N.m
what is the relationship between turtuosity and diffusion?
Answers
Construct the plane-stress yield envelopes in a principle stress space for both the Tresca and the von Mises yield theories using your calculated value of the yield strength to scale the envelopes. Indicate the two equivalent load paths corresponding to pure shear on the yield envelopes. Calculate the shear yield strength of Al 6061-T6 aluminum predicted by the above theories.
Answers
Answer:
Explanation:
The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.
From the given information:
The elastic modulus is:
[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]
Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.
The shear yield strength by using von Mises criteria is estimated as;
[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]
The shear yield strength by using Tresca criteria is:
[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]
someone should help me and explain stress strain curve
Answers
Answer:
In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).
Explanation:
please mark brainliest
A shaft is made from a tube, the ratio of the inside diameter to the outside diameter is 0.6. The material must not experience a shear stress greater than 500KPa. The shaft must transmit 1.5MW of mechanical power at 1500 revolution per minute. Calculate the shaft diameter
Answers
Answer:
shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm
Explanation:
Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6
Shear stress of material ( Z ) ≤ 500 KPa
power transmitted by shaft ( P ) = 1.5MW of mechanical power
Revolution ( N ) = 1500 rev/min
Calculate shaft Diameter
Given that: P = [tex]\frac{2\pi NT}{60}[/tex] ---- 1
therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 ) = 9.554 KN-M
next
[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]
hence ; T = Z[tex]_{p} *Z[/tex]
attached below is the remaining part of the solution
what's nested piezometer?
Answers
Explanation:
Nested piezometers indicate an upward flow if the elevation of the top of the water in the piezometer tube that penetrates the aquifer to the deeper point is greater than the elevation of the water in the shallower tube
In Databrawl, what team is most powerful?
A) Virus
B) Malware
C) Firewall security
D) Programs
Answers
Answer:
I would say Firewall security. Malware is pretty corrupting, but the firewall I think is good for protection if you have a good one.
Explanation:
electrical engineering
Answers
Answer:
Electrical engineering is an engineering discipline concerned with the study, design and application of equipment, devices and systems which use electricity, electronics, and electromagnetism.
Explanation:
QUESTION 6
Which of the following is NOT a resume format?
01. Chronological
O2. Portfolio
3. Functional
04. Combination
Answers
A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 22,000 N (4950 lbf), compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0
Answers
Answer:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
attached below is a detailed solution
Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.
Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
b. Would the heat rates for both arrangements change significantly if the air velocities were increased by a factor of 2?
c. What is the air mass rate requirement for the slotted nozzle arrangement?
Answers
Answer:
a. 2.30
b. decreases with increasing velocity.
c. 0.179 kg/s.
Explanation:
Without mincing let's dive straight into the solution to the question above.
[a].
The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:
The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.
While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.
Array of slot noozle = [10 × (2 × 0.004)]/ 20.92 × 10^-6] = 3824.
where A = 4/56 =0.714.
And Ar = [ 60 + 4 (40/2 × 4) - 2 ]^2 ]-1/2 = 0.1021.
N = 2/3 (0.1021)^3/4 [ 2 × 3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.
h = 24.3 × 0.030/0.004 = 91.1 W/m^2k.
Therefore; 659.6 × 0.030/0.5 = 39.0 W/m²k.
The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.
The slot noozle = 91.1 × 0.5 × 0.5 [ 140 -15] = 2846.87W.
The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.
[b].
2.3 [ (2^2/3)/ 2^4/5] = 2.1
Thus, it decreases with increasing velocity
[c].
The air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in your wire supply cabinet, you find a cardboard tube with single-conductor wire wrapped uniformly around it to form a solenoid. You carefully count the turns of wire and find that there are 570 turns. The diameter of the tube is 8.10 cm, and the length of the wire-wrapped portion is 35.0 cm. You pull out your calculator to determine the following.
a. the inductance of the coil (in mH)
b. the emf generated in the coil if the current in the wire increases at the rate of 3.00 A/s (Enter the magnitude in mV.)
Answers
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
A 5-m-long, 4-m-high tank contains 2.5-m-deep water when not in motion and is open to the atmosphere through a vent in the middle. The tank is now accelerated to the right on a level surface at 2 m/s2. Determine the maximum gage pressure in the tank. Mark that point at the interior bottom of the tank. Draw the free surface at this acceleration.
Answers
Answer: hello your question lacks the required diagram attached below is the diagram
answer : 29528.1 N/m^2
Explanation:
Given data :
dimensions of tank :
Length = 5-m
Width = 4-m
Depth = 2.5-m
acceleration of tank = 2m/s^2
Determine the maximum gage pressure in the tank
Pa ( pressure at point A ) = s*g*h1
= 10^3 * 9.81 * 3.01
= 29528.1 N/m^2
attached below is the remaining part of the solution
Leland wants to work in a Production career operating heavy machinery. Which type of education or training should Leland seek?
a bachelor’s degree then a master’s degree
vocational school certificate or master’s degree
on-the-job training or vocational school certificate
associate’s degree then a bachelor’s degree
Answers
Answer:
it is indeed C
Explanation:
Answer:
c
Explanation:
What two units of measurement are used to classify engine sizes?