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gavmur [86]
3 years ago
13

If a psychologist were to run an experiment measuring the effects of temperature on aggression the differing temperature would b

e considered the 
       A. dependent variable.   B. experimental variable.   C. independent variable.   D. random assignmen
Physics
1 answer:
Yanka [14]3 years ago
4 0

Answer:

(C) Independent variable

Explanation:

Independent is the variable used by the experimenter to vary a condition of interest in the experiment (temperature in this case). This measure  is varied over a sensible range to induce changes in other aspects (dependent variables; level of aggression in this case) which are measured during the experiment.

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A potter’s wheel of radius 17 cm starts from rest and rotates with constant angular acceleration until at the end of 32 s it is
dedylja [7]

Answer:

α=0.625rad/s^2

v=340m/s

w=10rad/s

θ=320rad

Explanation:

Constant angular acceleration = ∆w/∆t

angular acceleration = 20/32

α=0.625rad/s^2

Linear velocity v=wr

v = 20×17= 340m/s

Average angular velocity

w0+w1/2

w= 0+20/2

w= 20/2

w=10rad/s

What angle did it rotate with

θ=wt

θ= 10×32

=320rad

4 0
3 years ago
An electric current of 200 A is passed through a stainless steel wire having a radius R of 0.001268 m. The wire is L = 0.91 m lo
denis23 [38]

Answer:

The value is   T_m  =  435.2 \  K

Explanation:

From the question we are told that  

   The  current is  I  =  200 \ A

   The radius is  R =  0.001268 \  m

   The  length of the wire is  L  =  0.91 \  m\

    The  resistance is  R  =  0.126 \  \Omega

    The  outer surface temperature is  T _o  =  422.1 \  K

    The average thermal conductivity is  \sigma  =  22.5 W/mK

   

Generally the heat generated in the stainless steel wire is mathematically represented as  

    Q =  \frac{Power}{ \pi r^2L}

     Q =  \frac{I^2 R}{ \pi r^2L}

=>   Q =  \frac{200^2 * 0.126}{3.142 *  (0.001268)^2 * 0.91}

=>   Q =  1.096*10^{9}\  W/m^3

Generally the middle temperature is mathematically represented as

      T_m  =  T_o  + \frac{Q * r^2 }{ 6  * \sigma }

       T_m  =  422.1  +  \frac{1.096*10^{-9} * 0.001268^2}{6 * 22.5}

       T_m  =  435.2 \  K

4 0
3 years ago
In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used. The two thermocouples in each sample are
sergeinik [125]

Answer:

K = .3941 × 10³ W/m.K

Explanation:

Qcond = K A ΔT÷ L

∴K = Qcond ×L ÷ A ΔT

J ÷ S = P

P = I × V =Qcond

∴Qcond = I × V

               = 0.6 A × 110 V

               =66 W

L = 0.12 m

ΔT = 8 °C

Qcond =33 V

Area = (πD²) ÷ 4

       = [π (4 × 10⁻² )²] ÷  4

        = 1.256 × 10⁻³ m²

∴A = 1.256 × 10 ⁻³³ m²

So K = ( Qcond × L ) ÷ A ΔT

         = (33) (0.12 ) ÷ (1.256 ×10⁻³ ) × 8

         = 0.3941 × 10³ W/m .K

7 0
3 years ago
Chang sees the following formula on a website about Newton’s second law.
yarga [219]
The correct answer is the correct answer is A
3 0
3 years ago
Read 2 more answers
Please need help on this
Rasek [7]

Im pretty sure this is A. i may be incorrect, give brainliest if im correct please!

7 0
4 years ago
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