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Sati [7]
3 years ago
11

In your own words, what was the purpose of the field study?

Physics
1 answer:
DanielleElmas [232]3 years ago
6 0

Answer:

Hi there! I have only the procedure and the scientific names of the creatures. Hope this helps!

Procedure

I studied the physical features of ten creatures and classified them using the key. I completed each section below and recorded the creature's scientific name in the data section.

The Creatures Scientific Names:

1  

Fuzzus tallywag

2  

Fuzzus pointilus

3  

Silkus duosquirmus

4  

Fuzzus chompilus

5  

Silkus stretchilus

6  

Silkus tallyhas

7  

Fuzzus feelzalot

8  

Silkus monosquirmus

9  

Fuzzus squarilus

10  

Silkus monowrestle

Have a thrilling Thursday!

~Lola

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meriva
A Framework for K–12 Science Education: Practices, Crosscutting Concepts, and Core Ideas (Framework) recommends science education in grades K–12 be built around three major dimensions: science and engineering practices, crosscutting concepts that unify the study of science and engineering through their common application across fields, and core ideas in the major disciplines of natural science.
4 0
3 years ago
Assume your mass is 60 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb
Andre45 [30]

Answer:

W=1705.2 J

Explanation:

Given that

mass ,m= 60 kg

Acceleration due to gravity ,g= 9.8 m/s²

Height ,h= 2.9 m

As we know that work done by a force given as

W = F . d

F=force

d=Displacement

W=work done by force

Now by putting the values

F= m g (Acting downward  )

d= h  (Upward)

W= m g h    ( work done against the force)

W= 60 x 9.8 x 2.9 J

W=1705.2 J

Therefore the answer will be 1705.2 J.

8 0
3 years ago
What is newton's 3rd law of physics ​
ValentinkaMS [17]

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

4 0
3 years ago
If two charged objects in a laboratory are brought to a distance of 0.22 meters away from each other. What is
zysi [14]

Answer:

q_2=2.47\times 10^{-4}\ C

Explanation:

The charge on one object, q_1=9.9\times 10^{-5}\ C

The distance between the charges, r = 0.22 m

The force between the charges, F = 4,550 N

Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :

F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{4550\times (0.22) ^2}{9\times 10^9\times 9.9\times 10^{-5}}\\\\q_2=2.47\times 10^{-4}\ C

So, the charge on the other sphere is 2.47\times 10^{-4}\ C.

7 0
3 years ago
Whats the force of gravitation of a 10kg rock and 100kg boulder which are 5 meters apart​
spayn [35]

Answer:

F = 2.6692 x 10⁻⁹ N

Explanation:

Given,

The mass of the rock, m = 10 kg

The mass of the boulder, M = 100 kg

The distance between them, d = 5 m

The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula

                                   <em> F = GMm/d²  newton</em>

Where,

                                 G - Universal gravitational constant

Substituting the given values,

                                 F = 6.673 x 10⁻¹¹ x 100 x 10 / 5²

                                 F = 2.6692 X 10⁻⁹ N

Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N

6 0
3 years ago
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