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Triss [41]
3 years ago
9

Charles is having a lot of problems with errors in a very complicated spreadsheet that he inherited from a colleague, and he tur

ns to another co-worker, Seymour, for tips on how to trace errors in the sheet. If Charles sees which of the following, Seymour explains, there is a mistyped function name in the sheet.
a.#FORM?
b.#NAME?
c.#####
d.#FNCT?
Physics
1 answer:
grin007 [14]3 years ago
3 0

Answer:

<u>b.#NAME?</u>

Explanation:

Remember, in Spreadsheet programs like Ms Excel several types of errors can occur such as value error.

However, since Seymour explains that there is a mistyped function name in the sheet it is more likely to display on the affected cell as #NAME?.

For example the function =SUM is wrongly spelled =SOM.

Therefore it is important to make sure the function name is spelled correctly.

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Which statement is true?
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A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

M=-\frac{74}{2.4}

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0
4 years ago
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