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3 years ago

Computer programs like spreadsheets can be used to help organize and analyze data. True or false

Physics

2 answers:

satela [25.4K]3 years ago

7 0

Answer:

The statement is TRUE

Explanation:

computer programs like spreadsheets can be used to help organize and analyze data. Regression Analysis can be carried out in Excel to determine the relationship between two quantitative variables.

snow_lady [41]3 years ago

6 0

Computer programs like spreadsheets are used to organize and analyze data is a TRUE statement.

Explanation:

The spreadsheets or the database access programs of computer are the ones which helps us to organize a set of data significantly and thus helps to analyze them easily. Taking instance of spreadsheet, each data represented by the user defined parameter and set of value which represents the parameter.

Each of these are well input in single cell and the cell are spread over sheet in form of rows and column. This helps to give a good organized set of data representation and thus helps in analysis easily.

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One of the possible criteria of Schizophrenia is hallucinations.

Helen [10]

It is true but that’s not Physics

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1 year ago

The first asteroid to be discovered is Ceres. It is the largest and most massive asteroid in our solar system’s asteroid belt, h

stira [4]

Answer:

$4.81*10^{29}J$

Explanation:

Since the formula for kinetic energy of an object is:

$E_k = \frac{mv^2}{2}$

Where m is the mass of the object and v is the speed. We can substitute m = $3*10^{21}$ kg and v = 17900m/s:

$E_k = \frac{3*10^{21} * (17900)^2}{2} = 4.81*10^{29}J$

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2 years ago

A battery has a potential resistance of 12 V and a current of 1280 mA. What is the resistance? PLS HELP ME ASAPPPPPPPPPP

Romashka-Z-Leto [24]

Resistance (R) = 9.375 ohm (Ω)

Steps:

$R = \frac{V}{I}$

$= \frac{12 \: volt}{1280 \: milliampere}$

$= \frac{12 \: volt}{1.28 \: ampare}$

$= 9.375 \: ohm (Ω)$

6 0

2 years ago

A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its

Wittaler [7]

The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.

4 0

3 years ago

An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.

I am Lyosha [343]

Answer:

The density of this object is approximately $1.36\; {\rm kg \cdot L^{-1}}$ .

The density of the oil in this question is approximately $0.600\; {\rm kg \cdot L^{-1}}$ .

(Assumption: the gravitational field strength is $g =9.806\; {\rm N \cdot kg^{-1}}$ )

Explanation:

When the gravitational field strength is $g$ , the weight $(\text{weight})$ of an object of mass $m$ would be $m\, g$ .

Conversely, if the weight of an object is $(\text{weight})$ in a gravitational field of strength $g$ , the mass $m$ of that object would be $m = (\text{weight}) / g$ .

Assuming that $g =9.806\; {\rm N \cdot kg^{-1}}$ . The mass of this $63.8\; {\rm N}$ -object would be:

$\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}$ .

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was $(63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}$ . Hence, the weight of water that this object displaced would be $47.0 \; {\rm N}$ .

The mass of water displaced would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}$ .

The volume of that much water (which this object had displaced) would be:

$\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}$ .

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately $4.793\; {\rm L}$ .

The mass of this object is $6.50\; {\rm kg}$ . Hence, the density of this object would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately $4.793\; {\rm L}$ .

The weight of oil displaced would be equal to the magnitude of the buoyancy force: $63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}$ .

The mass of that much oil would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}$ .

Hence, the density of the oil in this question would be:

$\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}$ .

(Rounded to $\text{$3$ sig. fig.}$ )

7 0

2 years ago