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hodyreva [135]
3 years ago
5

Give me this same text but if it was a 7th grader

Physics
2 answers:
-Dominant- [34]3 years ago
7 0

Answer:

This should help!

What is 7th Grade Writing?

In Grade 7, students refine and build upon previously learned knowledge and skills in increasingly complex essays. On a regular basis, 7th grade students are expected to produce clear, coherent, and focused essays that are error-free. Seventh grade students are able to select and use different forms of writing for specific purposes such as to inform, persuade, or entertain. Students vary sentence structure and use verb tenses appropriately and consistently such as present, past, future, perfect, and progressive. Seventh grade students edit their writing based on their knowledge of grammar and usage, spelling, punctuation, and other conventions of written language. Seventh-graders use every phase of the writing process and continue to build their knowledge of writing conventions. Students draw data from multiple primary and secondary sources for use in research reports and projects.

The following writing standards represent what states* typically specify as 7th grade benchmarks in writing proficiency:

Grade 7: Writing Process

Seventh grade writing standards focus on the writing process as the primary tool to help students become independent writers. In Grade 7, students are taught to use each phase of the process as follows:

Prewriting: In grade 7, students generate ideas from multiple sources and use organizational strategies and tools such as technology, graphic organizers, notes, and outlines. Students choose the form of writing that best suits the intended purpose and then make a plan for writing that prioritizes ideas, addresses purpose, audience, main idea, and logical sequence.

Drafting: In seventh grade, students develop drafts by categorizing ideas, organizing them into paragraphs, and blending paragraphs within larger units of text. Writing exhibits the students’ awareness of the audience and purpose. Essays contain formal introductions, ample supporting details (e.g., facts, statistics, examples, anecdotes), and conclusions. Students analyze language techniques of professional authors, including concrete and abstract word choices, and infusing a variety of language techniques to reinforce voice.

Revising: In seventh grade, students revise selected drafts by elaborating, deleting, combining, and rearranging text. Goals for revision include improving coherence, progression, and the logical support of ideas and content. Grade 7 revision techniques include adding transitional words between sentences to unify important ideas and creating interest by using a variety of sentence structures (including the use of participles and participial phrases at the beginning and end of sentences). Students also evaluate drafts for voice, point of view, and precision of vocabulary. Seventh-graders use creative language devices, and modify word choices using resources and reference materials (e.g., dictionary, thesaurus).

Editing: Students edit their writing to ensure standard usage, varied sentence structure, and appropriate word choice (e.g., eliminating slang). Seventh-graders proofread for grammar, punctuation, capitalization, and spelling, using reference materials, word processors, and other resources.

Publishing: Using technology, seventh-graders refine and “publish” their work frequently in a format appropriate to the audience and purpose (e.g., manuscript, multimedia). Published pieces use appropriate formatting and graphics (e.g., tables, drawings, charts, graphs) when applicable to enhance the appearance of the document.

<h2>pls brailiest!</h2>

jarptica [38.1K]3 years ago
4 0

Answer:

Mount Rushmore is a very well built sculpture. It is visited by millions of people each year to see 60-foot tall images of the American presidents; George Washington, Thomas Jefferson, Abraham Lincoln and Theodore Rosevelt which are carved into the Black Hills of South Dakota. Mount Rushmore has faced many controversies too. The land it was built on is sacred to Native Americans. The Sioux were also forced to leave the land, after lots of battles

Explanation:

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What is the velocity of a 30-kg box with a kinetic energy of 6,000 J? 64 m/s
Nikolay [14]

Answer: 20 m/s

Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:

Ekinetic=(1/2)*(m*v^2)

then E=0.5*30Kg*(20 m/s)^2=15*400=6000J

8 0
3 years ago
Una grúa eleva un tubo de concreto de
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Explanation:

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7 0
2 years ago
The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
At 4.00 l, an expandable vessel contains 0.864 mol of oxygen gas. how many liters of oxygen gas must be added at constant temper
svet-max [94.6K]

Givens
=====
V = 4.00 L
T = 273oK We're assuming the temperature does not change, just the pressure.
n = 0.864 moles
R = 8.314 joules / mole  * oK
P = ?????

Formula
======
PV = n*R*T
P = n*R*T/V
P = 0.864 * 8.314 * 273 / 4
P = 490 kpa


You have to add 1.6 – 0.864 = 0.736 moles of gas.


We have to assume that the temperature and pressure remain the same when we add the 0.736 moles of gas. We are now looking for the volume.


PV = n*R*T

<span> V = 0.736 * 8.314 * 273 / 490</span>

V = 3.41 L Remember this is at about 4 atmospheres so we have to convert to Standard Pressure.

Total Volume = 3.41 + 4.00 = 4.41


V1 * P1 = V2 * P2

P1 = 490 kPa

P2 = 101 kPa

V1 = 7.41 L

V2 = ????


<span> <span> 7.41* 490 = V2 * 101 V2 = 7.41 * 490 / 101 V2 = 35.94 L </span> </span>


<span>You had 4 L now you need 31.94 more.</span> 
6 0
3 years ago
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