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denis23 [38]
3 years ago
9

PSYCHOLOGY! A significance error is an error that occurs from drawing an incorrect conclusion about the __________ within a stud

y.
A. critical value level
B. level of statistical support
C. number of participants
D. level of power

Physics
2 answers:
mars1129 [50]3 years ago
7 0

Hey!

-----------------------------------------------

Answer:

B. level of statistical support

-----------------------------------------------

Explanation:

When you create an incorrect conclusion within a study typically means that the study had something to do with a hypothesis. Well, if you have a statistical support and you have error in it, then the the lab is incorrect. One little thing can get someone confused.

-----------------------------------------------

Hope This Helped! Good Luck!

Dvinal [7]3 years ago
3 0

Answer:

B. level of statistical support.

Explanation:

A significance error is an error that occurs from drawing an incorrect conclusion about the level of statistical support.

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5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc
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Answer:

Δθ₁ =  172.5 rev

Δθ₁h =  43.1 rev

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       \Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2}  (1)  

  • Now, we need first to find the value of  the angular acceleration, that we can get from the following expression:

       \omega_{f1}  = \omega_{o} + \alpha * \Delta t  (2)

  • Since the machine starts from rest, ω₀ = 0.
  • We know the value of ωf₁ (the operating speed) in rev/min.
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       3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)

  • Replacing by the givens in (2):

       57.5 rev/sec = 0 + \alpha * 6 s  (4)

  • Solving for α:

       \alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)

  • Replacing (5) and Δt in (1), we get:

       \Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev  (6)

  • in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

       \Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev  (7)

  • In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

       \Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2}  (8)

  • First of all, we need to find the value of the angular acceleration during the second period.
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       \alpha_{2}  =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)

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