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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 list of factors best supports the concept that they are "just right" for the planet earth to support life?
Ivanshal [37]
Temperature, gravity, atmosphere and water
8 0
3 years ago
Read 2 more answers
How many miles can you get on one tank of gas if your tank holds 18 gallons and you get 22 miles per
MrRissso [65]
Multiply the two. The units will cancel out. 22 miles per gallon can be expressed as 22 miles / 1 gallon, so you can write the following:
18 gallons * 22 miles / 1 gallon
Which simplifies to 18 * 22 miles = 396 miles.
4 0
4 years ago
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A bicycle wheel with a radius of 0.42 m accelerates uniformly for 6.8 s from an initial angular velocity of 5.5 rad/s to a final
Art [367]

Answer:

The angular acceleration required  is 0.1765 rad/ s^2

Explanation:

The radius of the bicycle wheel has a radius of 0.42 m.

The acceleration is for time, t =  6.8 seconds.

Initial angular velocity is given as  \omega_{0}  = 5.5 rad/s

Final angular velocity is given as \omega_{f} = 6.7 rad/s

Therefore from the formula for angular speed we get

\omega_{f} = \omega_{0} + (\frac{d\omega}{dt} \times t),   where t is the time in seconds.

Therefore we get

6.7 =  5.5 + (6.8 × \frac{d\omega}{dt} )

Therefore we get the angular acceleration, \frac{d\omega}{dt} = \frac{(6.7 - 5.5 }{6.8}  = 0.1765 rad/ s^2

The angular acceleration required  is 0.1765 rad/ s^2

8 0
3 years ago
Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is
Elan Coil [88]

Answer:

The intensity of light from the 1mm from the central maximu is  I = 0.822I_o

Explanation:

From the question we are told that

                         The wavelength is \lambda = 620 nm = 620 *10^{-9}m

                         The width of the slit is w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m  

                          The distance from the screen is  D = 3.00m

                           The intensity at the central maximum is I_o

                          The distance from the central maximum is d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m

        Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

                     I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2

    Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

            sin \theta = \frac{z}{D}

           

if the angle between the the light at z and the central maximum is small

Then  sin \theta =  \theta

   Which implies that

              \theta = \frac{z}{D}

substituting this into the equation for the intensity

             I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D}  ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]

given that z =1mm = 1*10^{-3}m

   We have that

              I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2

                 =I_o [\frac{sin(0.760)}{0.760}] ^2

                 I = 0.822I_o

               

 

4 0
3 years ago
An object has a mass of 12 kg. on planet a, the object weighs 69 n. the force of gravity on planet a is:_____.
balandron [24]

the force of gravity on planet a is g = 5.72 m/s2

There is an object that weighs 12 kg. The object's weight on Planet A is 69 N. The equation F=mg, where F is the weight, m is the mass, and g is the force of gravity, can be used to calculate the force of gravity on Planet A:

g = F/m

g = 69 N/ 12 kg

g = 5.72 m/s2

<h3>What is the gravity force?</h3>

The force of attraction between any two objects in the universe is known as gravity or gravitational force. The mass of the object and the square of the distance between them determine the force of attraction. The weakest known force in nature, it is by far.

learn more about object has a mass here  

brainly.com/question/25959744

#SPJ4

8 0
2 years ago
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