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Marina CMI [18]
3 years ago
6

For which of these questions could a testable hypothesis be developed? Check all that apply. A. Are gases better than solids? B.

Do different liquids freeze at the same rate? C. Do cookies taste better when they are baked in a gas oven or in an electric oven? D. Does temperature affect the strength of a magnet? E. Does the color of water affect how quickly it evaporates?
Physics
2 answers:
Shalnov [3]3 years ago
7 0

Testable hypotheses could be developed for questions  <em>B,  D,  and  E</em> .

The main problem with  A  and  C  is the word "better".  This is a rubber word.  It means different things to different people, and it can mean different things to the same person at different times.  It's always a matter of opinion, it has no unique units of measure, and there is no concrete way to measure it.  Not good for developing a testable hypothesis.

OlgaM077 [116]3 years ago
3 0

Answer:

B, D, E

Explanation:

When determining if something could have a testable hypothesis you have to determine if you could test the hypothesis you come up with. An example hypothesis could be for B saying If I put different liquids in the freezer at the same time, then (liquid example) would freeze the fatest (or slowest) this is something you can test and can be measured by a test with data.

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A certain sea cow can paddle 2.0 m/s in still water. If she attempts to cross a river, from the south bank to the north with a c
aliina [53]

Answer:

v = 3.6m / s ,   θ = 56º

Explanation:

This is a relative speed exercise, let's use the Pythagorean theorem

        v = √ (v₁² + v₂²)

where v₁ is the speed of the sea still water and v₂ the speed of the current

         

let's calculate

       v = √ (2² + 3²)

        v = 3.6m / s

to find the direction we use trigonometry

      tan θ = v₂ / v₁

       θ = tan⁻¹ (v₂ / v₁)

let's calculate

       θ = tan⁻¹ (3/2)

        θ = 56º

4 0
3 years ago
A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied​
never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman  = 300N/m²

Area = 0.02m²

Unknown:

Force applied  = ?

Solution:

Pressure is the force per unit area on a body

        Pressure  = \frac{force}{area}

         Force  = Pressure x area

        Force  = 300 x 0.02  = 6N

8 0
3 years ago
A car's position in relation to time is plotted on the graph. What is the car's average velocity for segment C? A) -4 m/s B) -0.
kotegsom [21]

Answer:

b

Explanation:

its the anwser

7 0
3 years ago
Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a
Helen [10]

Complete Question

Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be added to the gas in the form of heat by applying a flame to the outside of the container. Conversely, energy can also be removed from the gas in the form of heat by immersing the container in ice water. Energy can be added to the system in the form of work by pushing the piston in, thereby compressing the gas. Conversely, if the gas pushes the piston out, thereby pushing some atmosphere aside, the internal energy of the gas is reduced by the amount of work done.

          pV=nRT

so the absolute temperature T is directly proportional to the product of the absolute pressure p and the volume V,Here n denotes the amount of gas moles,which is a constant because the gas is confined and R is the universal constant

What is the \triangle U as the system of ideal gas goes from point A to point B on the graph recall u is proportional to T

Answer:

\triangle T=0

\triangle V=0

The gas A and B have same internal energy

Explanation:

From the question we are told that

Pa=u atm\\Va=1m^3\\Pb=1 atm\\Vb=4m^3

Generally the equation of temperature is mathematically given as

Ta=\frac{Pv}{nR}

Ta=\frac{u*1}{nR}

And

Tb=\frac{PbVb}{nR}

Tb=\frac{u*1}{nR}

Generally the change in temperature \triangle T is mathematically given as

\triangle T=Tb-Ta=Tb=\frac{u*1}{nR}-\frac{u*1}{nR}

\triangle T=0

Generally the change in internal energy \triangle V

\triangle V=nC_v \triangle T\\

\triangle V=0

Therefore with

\triangle T=0

\triangle v=0

The gas A and B have same internal energy

6 0
3 years ago
Using only one management style with all people is the most effective leadership technique for any organization.
Softa [21]

the answer for this is false

4 0
3 years ago
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