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Alenkinab [10]
3 years ago
8

What is includeed in a cover sheet

Physics
1 answer:
Oksana_A [137]3 years ago
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>
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A comet fragment of mass 1.96 × 1013 kg is moving at 6.50 × 104 m/s when it crashes into Callisto, a moon of Jupiter. The mass o
vredina [299]

Answer:

Recoil speed, 1.17\times 10^{-5}\ m/s                          

Explanation:

Given that,

Mass of the comet fragment, m_1=1.96\times 10^{13}\ kg

Speed of the comet fragment, v_1=6.5\times 10^4\ m/s

Mass of Callisto, m_2=1.08\times 10^{23}\ kg

The collision is completely inelastic. Assuming for this calculation that Callisto's initial momentum is zero. So,

m_1v_1=(m_2+m_2)V

V is recoil speed of Callisto immediately after the collision.

V=\dfrac{m_1v_1}{(m_2+m_2)}\\\\V=\dfrac{1.96\times 10^{13}\times 6.5\times 10^4}{(1.96\times 10^{13}+1.08\times 10^{23})}\\\\V=1.17\times 10^{-5}\ m/s

So, the recoil speed of Callisto immediately after the collision is 1.17\times 10^{-5}\ m/s

7 0
4 years ago
Suppose you fill two rubber balloons with air, suspend both of them from the same point, and let them hang down on strings of eq
Ulleksa [173]

The charge on each the balloon is 100nC or 1.2 × 10^-7 C

Consider two balloons of diameter 0.200m each with a mass of 1.00g hanging apart with 0.0500m separation on the ends of string making angles of 10.0° with the vertical.

The charge on each balloon can be found from

F_{e} = \frac{k_{e}q^{2} }{r^{2}} \\q = \sqrt{\frac{k_{e}q^{2} }{r^{2}}} \\\\q = \sqrt{\frac{(2\times10^{-3N}(0.25m)^{2}}{8.99\times10^{9}N\cdotm^{2}/C^{2}}\\

q = 1.2\times10^{-7}C or 100nC

An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge. The matter is positively charged if it contains more protons than electrons and negatively charged if it contains more electrons than protons.

Learn more about the charge here:

brainly.com/question/14713274

#SPJ4

3 0
1 year ago
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if
sergij07 [2.7K]

Explanation:

a) I=\displaystyle \sum_{i}m_ir_i^2

where r_i is the distance of the mass m_i from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2

7 0
3 years ago
What is the frequency, in units of kiloHertz, of an AC waveform that has a period of 12 microseconds?
Alik [6]

Answer:

83.3 kHz

Explanation:

The frequency of a waveform is equal to the reciprocal of its period:

f=\frac{1}{T}

where

f is the frequency

T is the period

In this problem, we have

T=12 \mu s=12\cdot 10^{-6} s

so, the frequency of the waveform is

f=\frac{1}{12 \cdot 10^{-6} s}=8.33\cdot 10^4 Hz

And by converting into kiloHertz,

f=8.33\cdot 10^4 Hz=83.3 kHz

3 0
3 years ago
In 1947 Thor Heyerdahl sailed a simple raft from Peru to Polynesia, following the ocean currents for more than 6,000 kilometers.
GenaCL600 [577]
The correct answer of the given question above would be option C. In 1947 Thor Heyerdahl sailed a simple raft from Peru to Polynesia, following the ocean currents for more than 6,000 kilometers.<span> This statement accurately describes what Heyerdahl proved by this voyage. It would have been possible for people from ancient Peru to reach Polynesia by following ocean currents. </span>
7 0
3 years ago
Read 2 more answers
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