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3 years ago

Which of the following statements describe the steps you need to take to delete a file?

2 answers:

6 0

In order to delete a file, it is required to first select the file which you want to delete and then select 'delete' from the file menu. It will then ask you to confirm the deletion which will be required in order to complete the file deletion process.

It isn't required to click on the X at the top right of the screen or to open the file you want to delete as these steps aren't relevant to the deletion process.

Hence, the statements which describe the steps you need to take in order to delete a file are as follows:

- Confirm the deletion.
- Select Delete from the File menu.
- Select the file you want to delete.

olga55 [171]3 years ago

4 0

<h3>Explanation:</h3>

The process is as follows:

Select the file you want to delete.

Select delete from file menu. (You can right click on the file or press delete button on keyboard). A pop would appear which would prompt you to confirm the deletion of the file.

Confirm the deletion.

The x button at the top right would close the file and not delete it. Thus, it is incorrect option.

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What is tensile stress?

Rasek [7]

Tensile strength is the amount of tension a material can hold, at least I hope that’s what it is.

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4 years ago

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe

pentagon [3]

Answer:

(A) l = 0.39 m

(B) l =0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg(1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)

therefore we now have

mg(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l =0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 +l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0

3 years ago

What is relative density?

Artemon [7]

Relative density, or specific gravity, is the ratio of the density of a substance to the density of a given reference material. Specific gravity for liquids is nearly always measured with respect to water at its densest; for gases, the reference is air at room temperature.

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3 years ago

Which planet is smaller than Earth’s core?

makvit [3.9K]

Answer:

The answer <em><u>is C. Mars</u></em>. Mars and Mercury are both smaller than Earth's core. Hope this helps you :)

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3 years ago

Read 2 more answers

P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar

creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0

3 years ago