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2 years ago

The __________________ refers to the main screen of the computer.

Engineering

1 answer:

Sedbober [7]2 years ago

4 0

Answer:

Desktop

Explain:

Desktop refers to the main screen of the computer. It is the first screen you see after logging in. The desktop’s appearance can vary widely because it is highly customizable, but generally desktops will feature a large image, icons, and a taskbar(covered later on this page).

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Pipelines are a useful means of transporting oil because they: Multiple select question. are fast never fail to deliver are chea

Rufina [12.5K]

Pipelines are a useful means of transporting oil because they offer low maintenance and dependable transportation for a narrow but important range of products.

<h3>What is a pipeline?</h3>

A pipeline is a system of connected pipelines that can be either underground or out in the environment. These pipelines are used to transport or distribute water, gas, and oil.

The options are attached

a. Pipelines provide jobs for consumers because of the resurgence of exploration and drilling in North America.

b. Pipelines are versatile, carrying more ton-miles than any other mode of transport over more than 2 million miles of pipeline.

c. Pipelines have more locations than water carriers.

d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.

Thus, the correct option is d. Pipelines offer low maintenance and dependable transportation for a narrow but important range of products.

Learn more about Pipelines

brainly.com/question/14266025

#SPJ1

7 0

1 year ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

Explanation:

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

2 years ago

Select the correct answer.

cricket20 [7]

Answer:

The power generated by a wind farm is not constant because of irregular wind patterns.

5 0

2 years ago

Which of the following answer options are your employer's responsibility?

tino4ka555 [31]

Answer:

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Explanation:

To find - Which of the following answer options are your employer's responsibility? Select all that apply.

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Solution -

The correct options are -

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

All are the Responsibilities of an employer

Reason -

The most important duty of the employer is to stay alert and implement a correctly and efficiently written communication program related to hazards of the substances in the workplace.

He also has to maintain the program so that employees do not get affected.

3 0

2 years ago

Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a radius of gyration of 75 mm; gear C

Kruka [31]

Answer:

(4.5125 * 10^-3 kg.m^2)ω_A^2

Explanation:

solution:

Moments of inertia:

I = mk^2

Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2

Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2

Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2

Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.

r_A=50 mm

r_1 =100 mm

r_2 =50 mm

r_C=150 mm

At the contact point between gears A and B,

r_1*ω_b = r_A*ω_A

ω_b = r_A/r_1*ω_A

= 0.5ω_A

At the contact point between gear B and C.

At the contact point between gears A and B,

r_C*ω_C = r_2*ω_B

ω_C = r_2/r_C*ω_B

= 0.1667ω_A

kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2

=(4.5125 * 10^-3 kg.m^2)ω_A^2

6 0

2 years ago