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3 years ago

Essay on the problems kenge will face going to college in the story seeing is believing

Physics

1 answer:

Ilia_Sergeevich [38]3 years ago

6 0

Essay on the problems kenge will face going to college in the story seeing is believing is described below

Explanation:

1.An essay is really just a body of words meant to convey something. So, in your class, you used an essay to convey the pros and cons of love

2.In the admissions essays, you are using an essay to convey something about yourself. You will probably be asked to write about the reasons you want to enter a specific field or college, an event in your life that helped shape you, or other personal subjects. If it helps, rewrite these prompts into questions:

REAL examples

Why do you want to go to Harvard?
Why do you want to enter the medical field?
Are there any events in your life that affected your grades, but are not reflected in the rest of your application?

3.While the standard INTRO, BODY, CONCLUSION essay is widely used, it is not required. It is used because it is an effective way to convey information:.

INTRO: grabs reader's attention, tells the reader the main topic, gives necessary background information, and lists what is going to be expressed about the topic
BODY: one paragraph per idea about the main topic (eg 1 for pros, 1 for cons; OR 1 telling about an event in your life, 1 explaining how that event affected you, 1 giving an example of how you used what you learned in a new situation)
CONCLUSION: summarizes what you already told the reader, gives a broader context, and explains how the information is useful

4.To summarize: an essay is just a body of words that expresses an idea. There are many ways to write one, but having an introduction, paragraphs separated into ideas, and a conclusion helps make your idea easier to understand and more memorable.

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Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

consider the attached diagram below

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )$

$V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )$

Answer

$V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }$

$V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }$

8 0

3 years ago

Which is considered computer storage

Korolek [52]

Answer:

Second Option

Explanation:

The "hard drive" or the second option is one of the main components of storing information on a computer. You already have a hard drive built into your computer, or laptop when you buy it, and you can buy additional hard drives in the form of plugins that can store even more data if your original hard drive becomes full of data.

Hope this helps.

8 0

3 years ago

Read 2 more answers

An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to

kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

6 0

3 years ago

Read 2 more answers

Suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg.

ladessa [460]

The other 4 kg of mass may have departed the scene
of the fire, in the form of gases and smoke particles.

7 0

3 years ago

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

bearhunter [10]

Answer:

The mass of the banana is m and it is at height h.