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Romashka [77]
3 years ago
9

What are the activities involved in scientific method?

Physics
1 answer:
klemol [59]3 years ago
6 0
 Ask a Question Do Background Research Construct a Hypothesis Test Your Hypothesis by Doing an Experiment Analyze Your Dat…a and Draw a Conclusion Communicate Your Results
.
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When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the s
skad [1K]

Answer:

The value a = 1.1842 \ m/s^2

Explanation:

From the question we are told that

The weight of the sky diver is W =  968 \  N

The magnitude of the drag force is D_f  =  1085 \  N

The mass of the sky diver is m  =  98.8 \ kg

Generally the net force acting on the sky diver is mathematically represented as

F = m * a  =  D_f  - W

Hence the acceleration of the sky diver is mathematically represented as

a =  \frac{D_f - W}{ m}

=> a =  \frac{1085 - 968}{  98.8}

=> a = 1.1842 \ m/s^2

7 0
3 years ago
How does mineral growth occur?
marissa [1.9K]
Atoms are added to crystal faces.
3 0
3 years ago
Read 2 more answers
How do i calculate this?
Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0
3 years ago
Express newton pascal joule and watt in fundamental unit​
Nadya [2.5K]

i added an image. i hope it helped :)

3 0
3 years ago
An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
jenyasd209 [6]

Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

7 0
3 years ago
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