Ashworth Lesson 2: Careers for mechanics and technicians exam

Sonja [21]

Horace and John Dodge founded the Dodge Brothers Company in Detroit in 1900, and quickly found work manufacturing precision engine and chassis components for the city's growing number of automobile firms. Chief among these customers were the established Olds Motor Vehicle Company and the new Ford Motor Company.

6 0

3 years ago

Elijah wants to make some changes to a game that he is creating. Elijah wants to move the bleachers to the right, the coaches to

Ahat [919]

Answer:

the editor

Explanation:

the variable is usually a number, and the conditional loop has nothign to do with movement/ where something is located

7 0

3 years ago

Read 2 more answers

A 900 kg car is accelerated from a speed of 10 m/s to 30 m/s. An estimated heat loss of 20 BTU's occurs during the acceleration.

Strike441 [17]

Answer:

Work = 651,1011 kJ

Explanation:

Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

$E_{in}- E_{out}=E_{system,final}- E_{system,initial}$

Where

$E_{in}- E_{out}=(Q_{in}-Q_{out})_{heat}+(W_{in}-W_{out})_{work}+(Em_{in}-Em_{out})_{mass}$

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

$E_{in}- E_{out}=-Q_{out}+W_{in}$

Regarding the energy system we have the following:

$E_{system,final}- E_{system,initial}=(U_{f}-U_{i})_{internal}+(1/2m(V^2_{f}-V^2_{i}))_{kinetic}+(mg(h_{f}-h_{i}))_{potential}$

By doing the calculations we have:

$E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ$

Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.

Finally, the work needed to move the car under the required conditions is calculated as follows:

$W_{in}=Q_{out}+E_{system,final}- E_{system,initial}\\W_{in}=21,1011+630=651,1011kJ$

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.

4 0

3 years ago

A solid circular shaft has a uniform diameter of 5 cm and is 4 m long. At its midpoint 65 hp is delivered to the shaft by means

AlekseyPX

Answer:

A) τ_max = 59.139 x 10^(6) Pa

B) θ = 0.0228 rad.

Explanation:

A) In the left half of the shaft we have 25 hp which corresponds to a torque T1 given by;

P = Tω

Where P is power and ω is angular speed.

Power = 25 HP = 25 x 746 W = 18650W

ω = 200 rev/min = 200 x 0.10472 rad/s = 20.944 rad/s

P = T1•ω

T1 = P/ω = 18650/20.944

T1 = 890.47 N.m

Similarly, in the right half we have 40 hp corresponding to a torque T2

given by;

P = T2•ω

T2 = P/ω

Where P = 40 x 760 = 30,400W

T2 = 30400/20.944 = 1451.49 N.m

The maximum shearing stress consequently occurs in the outer fibers in the right half and is given by;

τ_max = Tρ/J

Where J is polar moment of inertia and has the formula ;J = πd⁴/32

d = 5cm = 0.05m

J = π(0.05)⁴/32 = 6.136 x 10^(-7) m⁴

ρ = 0.05/2 = 0.025m

T will be T2 = 1451.49 N.m

Thus,

τ_max = Tρ/J

τ_max = 1451.49 x 0.025/6.136 x 10^(-7)

τ_max = 59139022.94 N/m² = 59.139 x 10^(6) Pa

B) The angles of twist of the left and right ends relative to the center are, respectively, using θ = TL/GJ

G = 80 Gpa = 80 x 10^(9) Pa

θ1 = (890.47 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0363 rad

Similarly;

θ2 = (1451.49 x 2)/(80 x 10^(9) x 6.136 x 10^(-7)) = 0.0591 rad

Since θ1 and θ2 are in the same direction, the relative angle of twist between the two ends of the shaft is

θ = θ2 – θ1

θ = 0.0591 - 0.0363

θ = 0.0228 rad.

6 0

3 years ago

A weight-lifting athlete raises a mass of 160 kg through a vertical distance of 1.4 m. What force did

Over [174]

Answer:

1568N

2195.2J

Explanation:

Given parameters:

Mass of the weight = 160kg

Distance = 1.4m

Unknown:

Force applied to lift the weight = ?

Energy expended = ?

Solution:

The force applied in moving a body with a given mass through a distance is the weight;

Force applied = mg

Where m is the mass

g is the acceleration due to gravity

i. Applied force = 160 x 9.8 = 1568N

ii. The energy used to lift the weight is given as;

Energy = mgh

h is the vertical distance

Energy = 1568 x 1.4 = 2195.2J

8 0

3 years ago