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3 years ago

What is the name of the part that supports the headlight assembly?

1 answer:

8 0

A headlight assembly is made up of many parts that work together smoothly. One of these components is the headlight bracket. Sometimes known as “headlight mounting bracket,” it keeps the headlamp affixed to the header. The headlight bracket is made from sturdy material like plastic or steel.

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Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1

Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis = $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0

3 years ago

Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u

Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

4 0

2 years ago

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg

b) 504 kJ

c) 514 kJ

Explanation:

Given

-The mass of C_o2 = 1 kg

-The volume of the tank V_tank = 1 m^3

-The added energy E = 14 W

-The time of adding energy t = 10 s

-The increase in specific internal energy Δu = +10 kJ/kg

-The change in kinetic energy ΔKE = 0 and The change in potential energy

ΔPE =0

Required

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.

(c)The energy transferred by the heat transfer W in kJ and the direction of

the heat transfer.

Assumption

-Quasi-equilibrium process.

Solution

(a) The volume and the mass doesn't change then, the specific volume is constant.

v= V_tank/m ---> 1/1= 1 m^3/Kg

(b) The added work is defined by.

W =E * t ---> 14 x 10 x 3600 x 10^-3 = 504 kJ

(c) From the first law of thermodynamics.

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0

3 years ago

When does the vc-turbo engine use lower compression ratios?.

Veronika [31]

Answer:

Explanation:

The VC-T engine (for "variable compression, turbocharged") can adjust its compression ratio between 8:1 and 14:1 on the fly, offering high-compression efficiency under light loads and the low compression needed for turbocharged power under hard acceleration.

7 0

2 years ago

How do we find percentage error in measuring voltage across a resistor

Black_prince [1.1K]

Answer:

use the percentage error relation

Explanation:

The percentage error in anything is computed from ...

%error = ((measured value)/(accurate value) -1) × 100%

The difficulty with voltage measurements is that the "accurate value" may be hard to determine. It can be computed from the nominal values of circuit components, but there is no guarantee that the components actually have those values.

Likewise, the measuring device may have errors. It may or may not be calibrated against some standard, but even measurement standards have some range of possible error.

6 0

2 years ago

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