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Dvinal [7]
3 years ago
11

Observations must be _____. subjective objective biased deductive

Physics
2 answers:
Dmitrij [34]3 years ago
6 0
Objective
because it means<span> based on measurement, or reasoning free of bias.</span>
aivan3 [116]3 years ago
3 0

Answer: Objective

An observation is the outcome of the experimentation procedure. It is based upon the measurements and virtual evidences obtained while applying procedures. The observations should be objective that means they should be noted or recorded as such without any manipulation. An analyst should not make assumptions and the observation should be unbiased for a particular parameter because results will not be accurate.

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A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .​
kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+at\qquad\qquad [1]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

Solving [1] for a:

\displaystyle a=\frac{v_f-v_o}{t}

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

\displaystyle a=\frac{6.6-0}{6.5}

a = 1.015\ m/s^2

The distance is now calculated with [2]:

\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}

x = 21.45 m

The distance the car traveled is 21.45 m

6 0
3 years ago
FREE BRAINLEIAST FOR FIRST GOGOGOGO!!!!heererer
Kamila [148]

Answer:

ME PLS

Explanation:

6 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
WILL MARK AS BRAINLIEST
kumpel [21]
The answer is either C or D.
8 0
3 years ago
Two wires are made of the same material and have the same length but different radii. They are joined end-to- end and a potentia
Radda [10]

Answer:

Current

I think The choose (B)

B. Electric current

6 0
3 years ago
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