2. When performing an alignment, what action should be taken immediately after putting a vehicle on the rack?

A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2

IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length $l$ for the deformation, we use the following relation;

$l$ = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

$l$ = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

$l$ = 3161.025289 / 7240

$l$ = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0

3 years ago

Which traits are common in all four career pathways of the Information Technology field? Check all that apply.

Komok [63]

Answer:

Accuracy and attention to detail, Problem solving and critical thinking skills, Knowledge of programming language .

Explanation:

It is a technological area in which a person learns how to develop computer hardware, including PCs, laptops, tablets, processing, networking, and other hardware parts. Another field of study in IT is Management Information Systems (MIS).

The IT industry's career paths can be categorized equally in the two primary field’s hardware and software areas

In hardware, there is Production, maintenance, research and development, and strategic planning.

In software, there is manufacturing, development, programming, software testing, and maintenance and support under software.

Computer operations, administration of databases, sales / marketing and data centre management are connected areas.

7 0

2 years ago

How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?

nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

= 20.4kg · 9.81 $\frac{N}{kg}$ · 1.50m

= 35.316 $\frac{N}{m}$

= 35.316 W

4 0

2 years ago

The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k

sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0

3 years ago

What are the well-known effects of electricity

Sever21 [200]

Answer:

Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.

3 0

3 years ago

Read 2 more answers