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Andrei [34K]
3 years ago
10

After measuring a sample of individuals, a researcher finds that one participant's score is three times bigger than another part

icipant's score. This pair of measurements must come from a(n) ________ scale.
Physics
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

<u>Ratio</u> scale.

Explanation:

  • Ratio scale is a quantitative scale used for variable measurement which allows any researcher to compare the intervals of the measured parameter.
  • Ratio scale is the 4th level of measurement having a zero point or character of origin.
  • Ratio scale possesses most of the features of the other three variable measurement scale i.e ordinal, nominal and interval scales.
  • Nominal variables are used to name, or label a series of values.
  • Ordinal scales provide a sufficient amount of information about the order of choices, such as one would be able to understand from using a customer satisfaction survey, while the interval scales give us the order of values and also about the ability to quantify the difference between each one.
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A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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Answer:

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Explanation:

given,

Amount of heat absorbed = 45 kJ

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change in temperature = T₂ - T₁

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Heat\ capacity = \dfrac{Heat\ absorbed}{\Delta T}

Heat\ capacity = \dfrac{45\ kJ}{2.7}

Heat\ capacity = 16.6\ kJ/^0C

Heat capacity of the object is equal to 16.6 kJ/°C

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Explanation:

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