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lukranit [14]
3 years ago
14

The inclusion of the scientific method in psychology is so important because it

Physics
1 answer:
Gekata [30.6K]3 years ago
4 0

I know this question is from 2017 but I am doing this for anyone else that is looking for a answer. The answer should be C.) Allowed psychology to be considered a science.

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Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , mea
lapo4ka [179]

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

m*2*R + Iw = 0

w = -\frac{2mR}{I}

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3 years ago
As the frequency of a wave decreases the wavelength
alexandr402 [8]
As frequency decreases , the wavelength will increase !!

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3 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

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4 0
3 years ago
The typical unit for a period used with Kepler's third law is
ollegr [7]
Well, if you're using the law to work with periods of Earth satellites,
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orbits, or 'minutes' for the LEOs.

But if you're using it to work with periods of planets, asteroids, or
comets, then you'd be working in days or years.
6 0
3 years ago
Now review the difference between the temperatures on Earth and Mars. Also look at their distances from the sun.
aksik [14]

Answer: Use Question cove you can get it faster you can get the answer faster! ;) hope this helps ;) but yeah use that and answer is done right away

Explanation: HOPE THIS HELPSS!! ;))

7 0
2 years ago
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