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andriy [413]
3 years ago
8

What Member function places a new node at the end of the linked list?

Engineering
1 answer:
Nadusha1986 [10]3 years ago
3 0
<h2>˜”*°•.˜”*°• Question •°*”˜.•°*”˜ </h2>

<em>What Member function places a new node at the end of the linked list? </em>

<h2>˜”*°•.˜”*°• Answer •°*”˜.•°*”˜</h2>

appendNode

<h2>˜”*°•.˜”*°• Explanation •°*”˜.•°*”˜</h2>

The appendNode() member function places a new node at the end of the linked list. The appendNode() requires an integer representing the current data of the node.

<h2>˜”*°•.˜”*°• Details •°*”˜.•°*”˜</h2>

Subject: Coding (?)

Grade: College

Keywords: Function, linked list, appendNode, integer

Hope this helped. <3

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3 years ago
Training is a way for employers to provide ____ to enable employees to protect themselves and others from injuries.
Andreyy89

Answer:

Training is a way for employers to provide tools to enable employees to protect themselves and others from injuries.

Explanation:

5 0
3 years ago
Read 2 more answers
The alignment readings for the front of a vehicle are shown above. Camber and toe are within specification, caster is not. Techn
dlinn [17]

Answer:

B. B only

Given Information:

1. Camber and toe are within specification

2. Caster is not within specification

Technician A says that with the current settings, the left front tire tread may wear on the inside edge.

Technician B says that with the current settings, the vehicle may pull to the left

Explanation:

Lets discuss the effects of Camber, toe and caster misalignment

Effects of Camber and Toe misalignment:

Camber is the inward or outward tilt of the fron tires and is used to distribute load across the tread. Any misalignment causes uneven loading on the tires which results in tire wear on one edge.

The most common cause of tire wear on the inside edge is due to the camber misalignment which results in premature tire wear.

Another reason is of tire wear is vehicle’s toe. A slight misalignment of the toe reduces the life of the tire.

Since it is given that camber settings and toe settings are within specification therefore, tire tread wear on the inside edge cannot happen if camber and toe are within specification.

Technician A cannot be right.

Effects of Caster misalignment:

Whenever there is a misalignment of the castor then the vehicle will not be able to go in straight line rather it will pull to either left or right side. Caster misalignment also causes heavy or light steering depending upon the positive or negative misalignment of caster.

Since it is given that caster settings are not within specification therefore, the vehicle may pull to the left due to the caster misalignment.

Technician B must be right.

4 0
4 years ago
Will a simple circuit work without a switch? Explain why or why not.
Marina CMI [18]

yes, a simple circuit will work without a switch. If we don't have a switch in a circuit then whichever device you are using electricity for wouldn't be able to switch off and will be working continuously. An open/close switch is usually all that is needed to supply or disconnect power from the source to the circuit.So, the circuit will work but it will work continuously.

7 0
3 years ago
Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa a
Elena L [17]

Answer:

$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide  with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

$ S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1}) $

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

$ S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50}) $

$ S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888) $

$ S_2 - S_1 = 0.5864 - 0.6968 $

$ S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

6 0
3 years ago
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