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Maslowich
4 years ago
7

Scientific models can be used for a variety of different purposes. Which of the following is not a possible use for a scientific

model?. a. study of complex systems. b. study of obsolete systems. c. replace study of the larger system. d. answer scientific questions.
Physics
2 answers:
alex41 [277]4 years ago
4 0
Options a to c can be the reasons for scientific models.

But to primarily answer scientific questions,that would require an empirical and experimental approach and not use of models.

Though after getting the answers, models can be built to further explain the answers.

<span>d. answer scientific questions.</span>
kvv77 [185]4 years ago
4 0
Scientific models can be used for a variety of different purposes. The one which<span> is not a possible use for a scientific model is:

</span>d. Answer scientific questions.

Explanation:

Scientific models are based on a set of observations. They can never be used for making predictions and are based on a set of observations. There are three main types of models, that are physical, mathematical, and conceptual. Models have<span> limitations but are useful and can be changed based on new evidence.
</span>A model<span> of the motions of the sun, moon, and earth can be considered as examples of scientific models.</span>
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4 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
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