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marusya05 [52]
2 years ago
11

After observing a moth that is camouflaged against dark-colored bark, a scientist asks a question. The scientist discovers that

the question has already been asked and answered by several investigations. What should the scientist do?The scientist should test the question anyway.
The scientist should broaden the scope of the question.
The scientist should come up with a new way to test the question.
The scientist should revise the question to address a gap in knowledge.
Physics
2 answers:
Rasek [7]2 years ago
6 0
Because of how it's worded the answer would most likely be number four                                                                                       

liq [111]2 years ago
6 0

Answer: The scientist should revise the question to address a gap in knowledge.

Explanation:

A research question is a scientific inquiry that is required to be answered using suitable methodology.

Among the options given, The scientist should revise the question to address a gap in knowledge. is the correct option. This is because of the fact that analyzing and finding the possibilities in the question again the scientist can add new facts in the question which have not been found out by the research conducted by the several scientists earlier. This will lead to the increment in the knowledge related to the research topic.

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Two blocks, 1 and 2, are connected by a rope R1 of negligible mass. A second rope R2, also of negligible mass, is tied to block
alekssr [168]

Answer:

Explanation:

Given

Two block are connected by rope R_1

R_2 rope is attached to block 2

suppose F_2 is a force applied to Rope R_2

Applied force F_2=Tension in Rope 2

F_2=(m_1+m_2)a---1

where a=acceleration of system

Tension in rope R_1 is denoted by F_1

F_1=m_1a---2

divide 1 and 2 we get

\frac{F_2}{F_1}=\frac{(m_1+m_2)a}{m_1a}

also m_1=2.11\cdot m_2

\frac{F_2}{F_1}=\frac{2.11m_2+m_2}{2.11m_2}

\frac{F_2}{F_1}=\frac{3.11}{2.11}

\frac{F_1}{F_2}=\frac{2.11}{3.11}

               

3 0
2 years ago
Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

7 0
2 years ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
Read 2 more answers
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit
Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle \theta_{a},  since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are n_{red} = 2.410 for red light and n_{blue} = 2.450 for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has n_{air} = 1.000.

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

\theta_{crit} = 24.09^{0}

Explanation:

Only the blue refracted ray is related to the critical angle in this question

n_{air} = 1.000

n_{blue} = 2.450

The relationship between the critical angle(\theta_{crit}), n_{air} and n_{blue} can be given as sin \theta_{crit} = \frac{n_{air} }{n_{blue} }

sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\  \theta_{crit} = 24.09^{0}

6 0
2 years ago
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