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levacccp [35]
3 years ago
8

Julian is late to class and sees a list on the board: -are explanations of observations and hypotheses -are based on many experi

ments -have been revised over time -can be used to predict what may be expected in the future what is julian’s class discussing?
Physics
2 answers:
Verizon [17]3 years ago
6 0

The four ways to incorporate evidence into the research paper are:

<span>1.    </span>If the research you are investigating has been with other research paper before.

<span>2.    </span>There is an experimental proof that the evidence is true.

<span>3.    </span>The research published belongs to a reliable source.

<span>4.    </span><span>There are citations of the evidence that you are investigating</span>

Roman55 [17]3 years ago
6 0

For edgenuty peeps its B. theories

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HELP!!
raketka [301]
Hello!

Since the two weights are <em>off</em> the table, the block will move towards letter F. 

I hope this helps :))
5 0
3 years ago
Read 2 more answers
What is the mass of 3 m3 of a substance having density 1200 kg/m3​
adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

4 0
2 years ago
What variable does each of the symbols in the equation represent?
taurus [48]

Answer:

Pls give the equation and for what is that equation for?

4 0
2 years ago
Topic Gravitational force amd firld strength.. help me please
I am Lyosha [343]

The gravitational force between <em>m₁</em> and <em>m₂</em> has magnitude

F_{1,2} = \dfrac{Gm_1m_2}{x^2}

while the gravitational force between <em>m₁</em> and <em>m₃</em> has magnitude

F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}

where <em>x</em> is measured in m.

The mass <em>m₁</em> is attracted to <em>m₂</em> in one direction, and attracted to <em>m₃</em> in the opposite direction such that <em>m₁</em> in equilibrium. So by Newton's second law, we have

F_{1,2} - F_{1,3} = 0

Solve for <em>x</em> :

\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

x \approx 6.74\,\mathrm m

5 0
3 years ago
Real bodies emit and absorb more radiation than a blackbody at the same temperature. True or False
monitta

Answer: False

Explanation:

Relative to the concept of radiations, a black body is an object capable of absorbing any form of electromagnetic radiation irrespective of its frequency or angle of incidence when incident on such object.

However, the same cannot be said about real bodies as real bodies are those which reflect all rays incident on them completely and uniformly in all directions.

One very important characteristic of black bodies is that they are ideal emmiters.

The concept of emmisivity is brought about by the existence of real bodies .

This is due to the fact that they are only able to emit radiation at a fraction of the black body energy levels.

Please note that by convention, the emmisivity of a real body is always less thaan 1.

As such they are not able to emit as much radiation as a black body at the same temperature.

3 0
3 years ago
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