3 years ago

In order to design an experiment you need a ____ about the scientific question you are trying to answer

Physics

2 answers:

s2008m [1.1K]3 years ago

6 0

C Hypothesis. It is the correct answer.

Nutka1998 [239]3 years ago

4 0

I think C:Hypothesis

You might be interested in

To raise an arm upward, the muscle on one side of it contracts. What happens to the muscle on the opposite side of the arm?

enyata [817]

Answer:

It relaxes

Explanation:

I took the test I hope it helps

5 0

3 years ago

Read 2 more answers

What is the resistance in a circuit that is connected to a 20 V battery and has a

Amanda [17]

Answer:

$R = \displaystyle\frac{V}{I}$

$R = \displaystyle\frac{20}{75}$

$R = 0.267$ Ω (approx)

7 0

3 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

A piece of wood and a piece of steel are at the same temperature; however, the steel feels hotter. This is because the steel has

Vikentia [17]

The steel traps the heat more making it hotter,you put this twice by the way.

3 0

3 years ago

A composite material is to be designed with epoxy (Em 3.5 GPa) and unidirectional fibers. The longitudinal elastic modulus of th

LenKa [72]

Answer:

Minimum elastic modulus of fiber = 455.64 GPa

Explanation:

Contents of composite material = Epoxy and Unidirectional fibers

Elastic modulus of epoxy = 3.5 GPa

Elastic modulus of composite material = 320 GPa

Volume fraction of fiber = 70 %

Volume fraction of epoxy = 100 - 70 = 30%

Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320

0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95

Elastic modulus of fiber = 455.64 GPa

Minimum elastic modulus of fiber = 455.64 GPa

5 0

3 years ago