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3 years ago

What is includeed in a cover sheet

1 answer:

4 0

Are you referring to try to get into a college? if you are here is a basic outlay...

Your Street Address
City, State, Zip Code

Date

Name of Person, Title
Company/Organization
Street Address
City, State, Zip Code

Dear Mr./Ms./Dr. :

Introduction: State your reason for writing. Name the specific position or type of work for which you are applying. (Mention how you heard about the opening, if appropriate.)

Body: Explain why you are interested in working for that employer, or in that field of work, and what your qualifications are. Highlight two to three achievements that relate to the position and field. Refer the reader to the enclosed resume, application, and/or portfolio.

Closing: Thank the reader for his or her time and consideration. Indicate your desire for an interview and provide your contact information. If the employer is willing to accept phone calls, state that you will call to discuss the possibility of scheduling an interview.

Sincerely,

Your Name
<span>Enclosure / Attachment
</span>

You might be interested in

How much work is done when 0.0050 c is moved through a potential difference of 9.0 v? use w = qv?

quester [9]

Answer:

0.045 J

Explanation:

The work done on a charge moving through a potential difference is given by

$W=q\Delta V$

where

W is the work done

q is the charge

$\Delta V$ is the potential difference

In this problem, we have

q = 0.0050 C is the charge

$\Delta V=9.0 V$ is the potential difference

Using the formula, we find the work done:

$W=(0.0050 C)(9.0 V)=0.045 J$

4 0

3 years ago

Read 2 more answers

A 120 V electric heater draws a current of 14.1 A. What is the resistance, in ohms?

Sav [38]

Answer:

8.5 ohms

Explanation:

from ohms law

V=IR

120=14.1R

divide both sides by 14.1

120/14.1=14.1R/14.1

R=8.5ohms

5 0

3 years ago

Read 2 more answers

A cell supplies current of 0.6A and 0.2A through 1ohms and 4.0ohms resistor respectively. Calculate the internal resistance of t

Vlad [161]

<h2>Answer:</h2>

0.5Ω

<h2>Explanation:</h2>

Since different currents are passing through the resistors, then the resistors are most probably connected in parallel. This also means that the same voltage will pass across them.

Using Ohm's law, the voltage across a resistor in a circuit is given by;

V = I(R + r) -----------(i)

<em>For the 1ohm resistor, the voltage across it is given by;</em>

<em>Where;</em>

I = current passing through the 1 ohm resistor = 0.6A

R = resistance of the 1 ohm resistor = 1Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.6(1 + r) -------------------(ii)

<em>For the 4.0ohm resistor, the voltage across it is given by;</em>

<em>Where;</em>

I = current passing through the 4.0 ohms resistor = 0.2A

R = resistance of the 4.0 ohms resistor = 4.0Ω

r = internal resistance of the cell = r

Substitute these values into equation (i) as follows;

V = 0.2(4.0 + r) -------------------(iii)

<em>Now solve equations (ii) and (iii) simultaneously;</em>

V = 0.6(1 + r)

V = 0.2(4.0 + r)

Substitute the value of V in equation (ii) into equation (iii). Therefore, we have;

0.6(1 + r) = 0.2(4.0 + r)

<em>Solve for r</em>

0.6 + 0.6r = 0.8 + 0.2r

0.6r - 0.2r = 0.8 - 0.6

0.4r = 0.2

r = $\frac{0.2}{0.4}$

r = 0.5

Therefore, the internal resistance of the cell is 0.5Ω

4 0

3 years ago

gas is compressed by a piston in a cylinder from an initial pressure of 1.0 bar and initial volume of 0.4m3 , to a final pressur

Ket [755]

The work was done on the gas. and internal energy of the gas will be -13459 J and -21649 J.

When energy is moved from one store to another, work is completed. Work done on the gas is taken as -ve.

Given data;

Initial pressure(P₁)=1.0 bar

The initial volume, V₁=0.4m³

Final pressure(P₂) =1.4 bar.

work is done on the gas., W=?(-ve)

Heat release= - 8.1 kJ

Change in the internal energy of the gas., ΔE

During the compression process, PV=C

P₁V₁=P₂V₂

1.0 bar × 0.4m³= 1.4 bar ×V₂

V₂=0.285 m³

$\rm W=P_1V_1ln\frac{V_2}{V_1} \\\\\ \rm W=1.0 \times 0.4 ln\frac{0.285}{0.4} \\\\\ W=-13549 \ J$

The work done on the gas will be -13459 J.

The internal energy is found as;

ΔE=q+w

ΔE= -8.1 ×10³ -13549

ΔE=-21649 J

Hence, work was done on the gas. and internal energy of the gas will be -13459 J and -21649 J.

To learn more about the work done by gas refer to;

brainly.com/question/12539457

#SPJ1

4 0

2 years ago

A mass m1= 6.20 kg is moving North with a velocity of v1= 13.5m/sec when it collides perpendicularly with another mass m2= 4.40

JulijaS [17]

Answer:

Momentum of mass $m_{1}$ is 83·7 kg m/s towards north

Momentum of mass $m_{2}$ is 38·72 kg m/s towards east

Explanation:

Given

Mass of $m_{1}$ = 6·2 kg

Mass of $m_{2}$ = 4·4 kg

Velocity of mass $m_{1}$ = $v_{1}$ = 13·5 m/s

Velocity of mass $m_{2}$ = $v_{2}$ = 8·8 m/s

Momentum of mass $m_{1}$ = $m_{1}$ × $v_{1}$ = 6·2 × 13·5 = 83·7 kg m/s

Momentum of mass $m_{2}$ = $m_{2}$ × $v_{2}$ = 4·4 × 8·8 = 38·72 kg m/s

As mass $m_{1}$ is moving North, the direction of momentum will also be in that direction because the direction of momentum will be in the direction of velocity

As mass $m_{2}$ is moving East, the direction of momentum will also be in that direction because the direction of momentum will be in the direction of velocity

∴ Momentum of mass $m_{1}$ is 83·7 kg m/s towards north

∴ Momentum of mass $m_{2}$ is 38·72 kg m/s towards east

7 0

3 years ago