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3 years ago

What allows for the investigation of a variable? Can someone help please.

Physics

2 answers:

ladessa [460]3 years ago

5 0

The answer is <span>C: Experiment </span>

nevsk [136]3 years ago

3 0

Answer: Experiment

Explanation: If you want to study a variable you should do small changes and observe how the variable changes.

This process is called an experiment, where the investigator does small changes and tries to understand how these changes affect the variable. Trying to create a model that describes the variable.

You might be interested in

1. Applied research observational evidence 2. Basic research the experimental factor that changes in response to a change in the

lesya [120]

Answer:

This question is about matching each definition with its correct term. Please find the term matched with their appropriate definition below.

Explanation:

1. Empirical evidence: An empirical evidence is an observational evidence i.e an evidence gathered by observation or use of senses.

2. Dependent variable: Dependent variable is an experimental factor that changes in response to a change in the independent variable. In other words, it is dependent on the independent variable.

3. Applied research: Applied research is a type of research oriented at solving a present problem or need. It encompasses the production of products that can be sold for profit.

4. Hypothesis: A hypothesis in an experiment is a proposed explanation for a scientific problem that itself can be tested by experimentation. A hypothesis aims at providing a testable explanation to an observed problem.

5. Control: A control is a quantity in an experiment that remains unchanged or constant. It is kept the same by the experimenter for all groups in the experiment in order not to influence the outcome.

6. Basic research: Basic research is the research that expands knowledge in a particular area. It is the kind of research that aims at filling a knowledge void or satiating curiosity.

7. Independent variable: The independent variable is the experimental factor that is changed or manipulated deliberately by the scientist.

8 0

3 years ago

A fan cart with the fan set to high rolled across the floor. The cart's speeds with the fan on high are shown below. If the fan

Vadim26 [7]

Answer:

Speed of cart's might be less than the high speed after 5 seconds.

Explanation:

Given that,

A fan cart with the fan set to high rolled across the floor.

Let the speed of fan cart with set to high is $x\ cm$ per second.

The fan supplies a force to the cart. If a lower fan speed were used, less force would be applied. This would cause a slower change in the cart's speed. So, the cart would be rolling more slowly than $x\ cm$ per second after 5 seconds. The speed of cart's might be less than $x\ cm$ per second.

Force is needed

A. for a moving object to keep moving at the same speed and direction

B. for a moving object to change its speed

C. for a motionless object to remain still

D. to prevent a moving object from turning

Hence,

Speed of cart's might be less than the high speed after 5 seconds.

3 0

2 years ago

A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha

Airida [17]

Answer:

The work made by the gas is 7475.69 joules
The heat absorbed is 7475.69 joules

Explanation:

We know that the differential work made by the gas its defined as:

$dW = P \ dv$

We can solve this by integration:

$\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv$

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

$P \ V = \ n \ R \ T$

$P = \frac{\ n \ R \ T}{V}$

This give us

$\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv$

As n, R and T are constants

$\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv$

$\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}$

$\Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 )$

$\Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})$

But the volume is:

$V = \frac{\ n \ R \ T}{P}$

$\Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )$

$\Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})$

Now, lets use the value from the problem.

The temperature its:

$T = 27 \° C = 300.15 \ K$

The ideal gas constant:

$R = 8.314 \frac{m^3 \ Pa}{K \ mol}$

So:

$\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm})$

$\Delta W = 7475.69 joules$

We know that, for an ideal gas, the energy is:

$E= c_v n R T$

where $c_v$ its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

$\Delta E = \Delta Q - \Delta W$

where $\Delta W$ is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

$\Delta E = 0$

$\Delta Q = \Delta W$

7 0

2 years ago

Simone is walking her dog on a leash. The dog is pulling with force of 34 N to the right and simone is pulling backward with a f

e-lub [12.9K]

I do not understand the full question, however if you are wondering which way Simone and the dog will go, they will go right because the force of 34 N from the dog is higher than the force of 16 N from Simone.

4 0

2 years ago

The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden

SSSSS [86.1K]

Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

let mass of water condensed = m

$E = m_1 L_f$