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3 years ago

A CUSTOMER BRINGS HER CAR INTO THE

2 answers:

7 0

A’ because at first you should know the condition of the headlight before then if they are in good condition you may find out from the fuse box.

In-s [12.5K]3 years ago

5 0

a bc if the bulbs are in a bad conditio. than u know that u dont have to remove it but only repair it.

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A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pr

Georgia [21]

Answer:

$W=-52 800\ \text{J}=-52.8\ \text{kJ}$

Explanation:

First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

$W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}$

6 0

3 years ago

Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact di

Ilya [14]

Answer:

Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.

5 0

3 years ago

Rafel knows that lessons learned is a valuable aid to future projects. When should he and his team address

Arada [10]

Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

5 0

3 years ago

A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi

saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, $\tau_{max}$ , is given by the following formula;

$\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )$

$t_w$ = 1 cm = 0.01

h = 29 cm = 0.29 m

$h_w$ = 25 cm = 0.25 m

b = 15 cm = 0.15 m

$I_c$ = The centroidal moment of inertia

$I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )$

$I_c$ = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

$I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}$

$I_c$ = 1.2257083 $\bar 3$ × 10⁻⁴ m⁴

From which we have;

$4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )$

Which gives;

W = 11,416.6879 N

$\sigma _{b.max} = \dfrac{M_c}{I_c}$

$\sigma _{b.max}$ = 1500 N/cm² = 15,000,000 N/m²

$M_c$ = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

$M_{max} = \dfrac{W \cdot L}{4}$

$L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417$

L ≈ 0.64417 m ≈ 64.417 cm.

4 0

3 years ago

Which of the following statements about a metal wire in equilibrium are true? Select all that apply. The electric field inside t

zlopas [31]

Answer:

The interior of the metal wire is neutral

The net electric field everywhere inside the wire is zero

There may be excess charges on the surface of the wire

There is no net flow of mobile electrons inside the wire

There are no excess charges in the interior of the wire

Explanation:

When a metal wire is in equilibrium, the interior of the metal wire is always neural and there is no net flow of mobile electrons inside the wire. Additionally, there are no excess charges in the interior of the wire but there is likelihood of having excess charges on the surface of the wire. Finally, the net electric field everywhere inside the wire is zero

6 0

3 years ago