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Westkost [7]
3 years ago
5

Mobo, a wireless phone carrier, completed its first year of operations on October 31. All of the year's entries have been record

ed, except for the following: a. At year-end, employees earned wages of $6,000, which will be paid on the next payroll date, November 6. b. At year-end, the company had earned interest revenue of $3,000. It will be collected December 1 -4 Part 3 Show the accounting equation effects of each required adjustment. (Enter any decreases to Assets, Liabilities, or Stockholders' Equity with a minus sign.) ansaction Assets Liabilities Stockholders' Equity 6,000 Salaries and Wages Payable 4,900 6,000 alaries and Wages Expense nterest Receivable a. 4.,900 Acounts Payable b. terest Revenue
Engineering
1 answer:
kenny6666 [7]3 years ago
3 0

Answer:

Explanation:

a) Salaries and wages expenses     $6,000

   To salary for wages payable        $6,000

b) Interest Receivable a/c-d-               $4,900

   To interest revenue                          $4900

Assets   = Liabiliteies +stockholder equity

a)       0   =                 6000    +    (-6000)

    (No effect)        (Increase)   (Decrease)

       0 =6000 -6000

b)  4900  = 0      +          4900

                 (no effect)    (increase)

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5 0
3 years ago
Translating type screws are used: A. In translating machines B. For imparting motion to a mechanical part C. As a fastener to co
ivanzaharov [21]

Answer:

(b) for imparting motion to a mechanical part

Explanation:

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4 0
4 years ago
Interpret the Blame responsibility and causation in your own words in the light of Columbia Accident.
Licemer1 [7]

Answer:

Proposed Improvements and Generic Lessons

Within 2 h of losing the signal from the returning spacecraft, NASA’s Administrator established the Columbia Accident Investigation Board (CAIB) to uncover the conditions that had produced the disaster and to draw inferences that would help the US space program to emerge stronger than before (CAIB, 2003). Seven months later, the CAIB released a detailed report that included its recommendations (Starbuck and Farjoun, 2005).

The CAIB (2003) report attempted to seek answers to the following four crucial questions:

1.

Why did NASA continue to launch spacecraft despite many years of known foam debris problems?

2.

Why did NASA managers conclude, despite the concerns of their engineers, that the foam debris strike was not a threat to the safety of the mission?

3.

How could NASA have forgotten the lessons of Challenger?

4.

What should NASA do to minimize the likelihood of such accidents in the future?

Although the CAIB’s comprehensive report raised important questions and offered answers to some of them, it also left many major questions unanswered (Starbuck and Farjoun, 2005).

1.

Why did NASA consistently ignore the recommendations of several review committees that called for changes in safety organization and practices?

2.

Did managerial actions and reorganization efforts that took place after the Challenger disaster contribute, both directly and indirectly, to the Columbia disaster?

3.

Why did NASA’s leadership fail to secure more stable funding and to shield NASA’s operations from external pressures?

By examining, with respect to the Columbia disaster, the case of NASA as an organization, one can try to extract generalizations that could be useful for other organizations, especially those engaged in high-risk activities—such as nuclear power plants, oil and gas, hospitals, airlines, armies, and pharmaceutical companies—and such generic principles may also be salutary for any kind of organization.

The CAIB (2003) report recommended developing a plan to inspect the condition of all RCC systems, the investigation having found the existing inspection techniques to be inadequate. RCC panels are installed on parts of the shuttle, including the wing leading edges and nose cap, to protect against the excessive temperatures of reentry. They also recommended that taking images of each shuttle while in orbit should be standard procedure as well as upgrading the imaging system to provide three angles of view of the shuttle, from liftoff to at least SRB separation. “The existing camera sites suffer from a variety of readiness, obsolescence, and urban encroachment problems.” The board offered this suggestion because NASA had had no images of the Columbia shuttle clear enough to determine the extent of the damage to the wing. They also recommended conducting inspections of the TPS, including tiles and RCC panels, and developing action plans for repairing the system. The report included 29 recommendations, 15 of which the board specified must be completed before the shuttle returned to flight status, and also made 27 “observations” (CAIB, 2005).

7 0
3 years ago
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna
Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

7 0
3 years ago
a resistor is made out of a long wire having a length L. Each end of the wire is attached to a termina of a battery providing a
Crank

Answer:

Option A

Solution:

As per the question:

Initially, voltage is V_{o}

Current is 'I' A

Length of the wire is L

Now,

We know that:

R = \frac{\rho L }{A}           (1)

where

\rho = resistivity\ of\ the wire

A = Cross sectional area of the wire

From eqn (1), if other things are taken to be constant, then

R ∝ L                 (2)

Thus

When the wire is cut into two reducing the length to \frac{L}{2}

Resistance, R' = \frac{R}{2}

Now, when these wires are connected as described, the connection is in parallel, therefore, the equivalent resistance of the two wires:

\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'}

\frac{1}{R_{eq}} = \frac{2}{R} + \frac{2}{R}

R_{eq} = \frac{R}{4}

Now, from Ohm's law:

I = \frac{V}{R}

Since, according to the question voltage V_{o} is constant, thus

I ∝ \frac{1}{R_{eq}}

I ∝ \frac{4}{R}

Thus

I becomes 4I

3 0
3 years ago
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