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Sergio039 [100]
2 years ago
13

What is the slope of the line plotted below?

Physics
1 answer:
Naddika [18.5K]2 years ago
3 0

Answer:

To find the slope, we need to have co-ordinates of atleast 2 points lying on the line.

In this case, the 2 points are

(-3,1) and (2,-2)

To find the gradient we put in the following rule :

Rise / Run = (y2 - y1) / (x2-x1) = (-2-1) / (2+3) = -3 /5

Explanation:

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Read the list of phrases from the article.
ICE Princess25 [194]

Answer:

(D)

to establish an understanding of key concepts relating to population biology

Explanation:

Thats what I would go with but I didn't read the article so I don't know what context was used. Good luck! :)

8 0
3 years ago
This diagram shows a heating curve for water.
Bess [88]

Answer:

it is B

Explanation:

Because I agree with her and also got 100

6 0
3 years ago
Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un
Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

          K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}

  \gamma = 1,     K_{IC} = 24.1

  [/tex]\sigma_{y}[/tex] = 570

and,   \sigma_{f} = 570 \times \frac{3}{4}

                       = 427.5

Hence, we will calculate the critical crack length as follows.

      a = \frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}

        = \frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}

       = 10.13 \times 10^{-4}

Therefore, largest size is as follows.

            Largest size = 2a

                                 = 2 \times 10.13 \times 10^{-4}

                                 = 20.26 \times 10^{-4}

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is 20.26 \times 10^{-4}.

4 0
3 years ago
On the ____ scale, an increase of ten dicibels means an intensity increases of ten times.
UkoKoshka [18]

Answer:

the answer is decibel

7 0
3 years ago
Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of
Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

A=2\times10^{-6}m^2

* The young modulus of the steel is,

Y=2.1\times10^{11}\text{ Pa}

Solution:

The young modulus of the steel in terms of the force and extension is,

Y=\frac{F\times l}{A\times dl}

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0
1 year ago
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