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3 years ago

Briefly explain why experiments having faulty design or inconsistent data are problems for scientists. List several reasons

Physics

1 answer:

Readme [11.4K]3 years ago

8 0

Answer:

There are several reasons that experiments with faulty designs or with inconsistent data are problematic for scientists. A person can make one of those problems if he or she were to poorly measure what they are studying. For example, someone measured the mass of a book correctly to be 2 pounds, and someone else measured it mistakenly to be 1 pound. Another way that a person can make problems with faulty designs and inconsistent data is the lack of accuracy and precision. This could happen when someone can have the value of 10 from a correct data set of 9, 10, 10, 11, and 12, and someone else can have the value of 10 from an incorrect data set of 5, 7, 19, 15, and 10. The first data set has a lot more precision that the second data set. Another example would be: Someone could have the value of 10 from a correct data set of 9, and 11. Someone else can have the value of 10, but have the incorrect data of 7, and 15. The first set has more accuracy than the second set. A third reason that faulty designed experiments and inconsistent data can happen is the flawed experiments. For flawed experiments to happen, they may be uncontrolled, untrustworthy conclusions, or being inconsistent with other tests performed. For the last reason that they can happen, there can be bias. this could happen when the samples are too small, not randomly selected samples, and some outliers are present.

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A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w

user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

$p\propto T$

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

where in this problem:

$p_1=0.981 atm$ is the initial pressure of the gas

$T_1=65^{\circ}+273=338 K$ is the initial absolute temperature of the gas

$T_2=41^{\circ}+273=314 K$ is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

$p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm$

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4 years ago

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3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

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1 kg air in a piston-cylinder assembly is heated at constant pressure, resulting the expansion of the volume. the initial temper

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Answer:

57,42 KJ

Explanation:

By a isobaric proces, the expresion for the works in the jpg adjunt. Then:

W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)

By the ideal gases law: PV=RTn

Then, in (1): (remember Pa = Pb)

W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)

Since we have 1 Kg air: How much is this in moles?

From bibliography: 28.96 g/mol

Then, in 1 Kg (1000 g) there are:

n = 34,53 mol

Finally, in (2):

W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ

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3 years ago

8. Noticing how much food is on your lunch<br> tray is a quantitative observation because

zhenek [66]

Answer:

you are making an observation that uses numbers.

Explanation:

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