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riadik2000 [5.3K]
3 years ago
8

Nalah labels ten rats with the numbers 1 through 10. Then, she records how long it takes each rat to run through a maze. She fin

ds it difficult to graph her data in a line graph. What change could she make to her experimental design that would allow her to plot a clearly displayed data trend in a single line graph?
A) Repeat the experiment ten times. Label the y-axis “Time” and the x-axis “Trial.” Plot the average time for all ten rats for each trial of the experiment.
B) Divide the rats into groups of females and males and repeat. Label the y-axis “Time” and the x-axis “Sex of Rat.” Plot the average time for each group.
C) Repeat the experiment ten times. Label the y-axis “Rat Number” and the x-axis “Time.” Plot the individual time for all ten rats for each trial of the experiment.
D) Divide the rats into two groups based on their speed from the initial data. Label the y-axis “Time” and the x-axis “Group.” Plot the average time for each group.

Answer: A) Repeat the experiment ten times. Label the y-axis “Time” and the x-axis “Trial.” Plot the average time for all ten rats for each trial of the experiment.
Physics
1 answer:
drek231 [11]3 years ago
5 0

Answer:

we have to make another question we maxed it out again lol

Explanation:

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IgorLugansk [536]

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Weddell seals make holes in sea ice so that they can swim down to forage on the ocean floor below. Measurements for one seal sho
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Answer:

B. 1 m/s

Explanation:

Metric unit conversions:

0.3 km = 300m

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v = s/t = 300/300 = 1m/s

So B is the correct answer

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If a guitar string vibrates 20 times each second,what is its frequency?
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The answer is 20. Frequency is "number of times" per unit time.
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What is the resultant of 5N force pointing north and 7N force pointing south? Do not forget include direction
balandron [24]

Given forceF

1

=5N and F

2

=7N and θ=60

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5

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+7

2

+2(5)(7)cos60

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25+49+35

F=

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4 0
3 years ago
A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric
Tcecarenko [31]

Answer:

a)   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m  and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s   c) v_{z}  = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

         Fe + Fm = m a

         a = (Fe + Fm) / m

the electric force is

         Fe = q E   k ^

         Fe = 4.25 10-4 60 k ^

         Fe = 2.55 10-2 k ^

the magnetic force is

         Fm = q v x B

         Fm = 4.25 10⁻⁴  \left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]

         fm = 4.25 10⁻⁴ (-j ^ 30 4)

         fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

      aₓ = 0

there is no force

Axis y

      ay = -5.10 10²/5 10⁻⁵ j ^

      ay = -1.02 107 j ^ / s2

z axis

      az = 2.55 10⁻² / 5 10⁻⁵ k ^

      az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

           x = vo t + ½ aₓ t₂2

           x = 30 0.02 + 0

           x = 0.6m

       

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

           y = I + go t + ½ ay t²

           y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

           y = 1 - 2.04 10³

           y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

          z = zo + v₀ t + ½ az t²

          z = 0 + 0 + ½ 5.1 10² 0.02²

          z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

   r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

 since there is no acceleration the speed remains constant

          vₓ = 30.0 m / s

Axis y

  let's use the equation v = v₀ + a_{y} t

         v_{y} = 0 + -1.02 10⁷ 0.02

          v_{y} = 2.04 10⁵ m / s

z axis

          v_{z} = vo + az t

          v_{z} = 0 + 5.1 10² 0.02

          v_{z}  = 1.02 10⁻¹m / s

8 0
4 years ago
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