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choli [55]
2 years ago
5

Which example is of researcher bias in interpreting data?

Physics
1 answer:
VikaD [51]2 years ago
5 0

Answer:

A researcher seeks to prove that babies smile at two months old. She counts some confused expressions as smiles.

Explanation:

This is the answer because its not actual date due to the fact she's using her opinion to decided whether or not the expression a baby is making is a smile, whereas the other answers had factual data involved.

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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later
Marta_Voda [28]
Velocity = displacement (distance)/time

v=80m/4s

v=20m/s

velocity = 20 meters per second
8 0
3 years ago
How do you write a hypothesis
Mashcka [7]
It's and if, then statement!
5 0
2 years ago
Answer the question based on this waveform.
Nuetrik [128]

Answer:

Cannot be determined from the given information

Explanation:

Given the following data;

Velocity = 24 m/s

Period = 3 seconds

To find the amplitude of the wave;

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.

However, the given parameters can be used to calculate the frequency and wavelength of the wave.

6 0
3 years ago
Would a measured force of (46.5 0.8 N  ) be in agreement with a theoretically calculated force of (48.4 0.6 N  ) ? Show your w
OverLord2011 [107]

Answer:

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )

Explanation:

From the question we are told that

  Measured force is  F_M  =  [46.5 \pm 0.8 \  N ]

   Calculated force is  F_c =  [48.4 \pm 0.6 \  N ]

Generally the measured force in interval form is

     46.5 - 0.8  < F_M  <  46.5 + 0.8

=>  45.7   < F_M  < 47.3

Generally the calculated  force in interval form is

     48.4 - 0.6  < F_c  <  48.4 + 0.6

=>  47.8   < F_M

Generally looking both interval we see that they do not intersect at any point Hence  

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )      

8 0
2 years ago
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