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vagabundo [1.1K]

3 years ago

6

What are two examples of observations you might make, and two examples of inferences you might make, when looking at logs burnin

g in a campfire?

1 answer:

Sedbober [7]3 years ago

5 0

Answer:

One observation would be the change in odor when observing the logs burning in a campfire and because of that a inference could be because a change in odor is occurring, a chemical change is happening to the logs (combustion).

A second observation would be that the wood being burned gave off smoke at first but then stopped and because of that a inference would be that the compound that were being burned creating the smoke was all evaporated from the wood.

You might be interested in

Is a step in the scientific method. The step that follows this step involves forming

svlad2 [7]

Answer: read this hope this helped

Explanation: A hypothesis is a possible explanation for a set of observations or an answer to a scientific question. ... The next step in the scientific method is to test the hypothesis by designing an experiment. This includes creating a list of materials and a procedure— a step-by-step explanation of how to conduct the experiment.

3 0

3 years ago

A 0.0427 kg racquet-ball is moving

Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball $(m_1)$ = 0.00427 kg

Velocity of racket-ball before collision $(v_{1i})$ = 22.3 m/s

Velocity of racket-ball after collision with box $(v_{1f})$ = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision $v_{2i}$ = 0 m/s

<em>[Since the box is stationary, so velocity is taken zero]</em>

Velocity of box moving forward after collision $v_{2f}$ = 1.53 m/s

To find the mas of the box $m_2$ .

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

$p_i=p_f$

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We can plugin the given value to find $m_2$

$(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)$

$0.9522+0=-0.4911+1.53m_2$

Adding both sides by 0.4911

$0.9522+0.4911=-0.4911+0.4911+1.53m_2$

$1.4433=1.53m_2$

Dividing both sides by 1.53.

$\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}$

$0.9433=m_2$

∴ $m_2=0.9433$ kg

Mass of the box = 0.9433 kg (Answer)

4 0

2 years ago

Which of the following accurately describes the way in which a muscle moves?

Vaselesa [24]

<h3><u>Answer</u>;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3><u>Explanation;</u></h3>

<em><u>The events of muscle fiber shortening occurs with in the sacromeres in the fibers. </u></em>
<em><u>Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.</u></em>
<em><u>Filament movement starts at the region or zone where thin and thick filaments overlap. </u></em>
<em><u>Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.</u></em>

7 0

2 years ago

Air is a good medium for sound waves because it is

Paul [167]

Answer:

air does not have a modulus of rigidity.

Explanation:

Since air is completely elastic medium, that is, it does not have a modulus of rigidity, therefore sound waves in air are longitudinal.

4 0

1 year ago

An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

3 0

3 years ago