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4 years ago

Why do you think scientists often find value in looking at and comparing anomalies within data??

2 answers:

7 0

Absolutely, as anomilies can show inconsistencies with data and hypothesis they may have and may actually assist in refining the accuracyof their ideas by not only encompassing the most commonly-occuring phenomena, but also the more seldom occuring scenarios as well.

Sonja [21]4 years ago

7 0

Answer:

Anomalies are important to verify the validity of the conclusions about experimients and can be the key to reinforce or disapprove a hypothesis. It can also contribute adding second observations about a single experiment.

Explanation:

Conducting experimients is a hard task by many reasons and one of them is collecting and interpreting data. If a scientist is not careful, He/She may bypass anomalies and get to misleading conclusions. Anomalies are important because allow us to question if the experiment was carried out properly and if its method can also lead to other observations than the ones stated by the scientist. Anomalies are essential to learn more about a single experiment and improve the way ivestigation is done professionally.

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Calculate the current in a circuit when the voltage of the circuit is 5 V and the circuit has a total resistance of 5 Ω.

seropon [69]

the answer is 1a as rearrange gives I = v divided by r

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3 years ago

A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effect

Fantom [35]

Answer:

$I_s=5.8A$

Explanation:

Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:

$P_p=P_s$

So:

$V_p*I_p=V_s*I_s$

Where:

$V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil$

Solving for $I_s$

$I_s=\frac{V_p*I_p}{V_s}$

Replacing the data provided:

$I_s=\frac{110*29}{550} =5.8A$

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3 years ago

De donde eres responde y te doy corona

irakobra [83]

soy de texas, united states

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3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$

Let us assume that, $A=0.0008\ m^2$

$\theta$ is the angle between normal and the magnetic field, $\theta=90^{\circ}-30^{\circ}=60^{\circ}$

Torque is given by :

$\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m$

So, the net torque about the vertical axis is $1.03\times 10^{-4}\ N-m$ . Hence, this is the required solution.

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3 years ago

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Katyanochek1 [597]

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