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Which of the following should be considered when analyzing the results of a scientific experiment? the hypothesis, but only if i

Zigmanuir [339]

<h2>Answer:</h2>

The correct answer is option C which is, "empirical evidence that was collected during the experiment".

<h3>Explanation:</h3>

Empirical evidence are the observations and data values collected during the experiment by using the senses.

Like if your are experimenting involving a chemical reaction, the temperate or color changes during the reaction should be counted in the interpretation of the results.

Hence the correct option is C.

6 0

3 years ago

Read 2 more answers

Two isolated copper plates, each of area 0.40 m2, carry opposite charges of magnitude 7.08 × 10-10 C. They are placed opposite e

polet [3.4K]

Answer:

The potential difference between the plates is 8 V.

Explanation:

Given that,

Area of plates = 0.40 m²

Charge $q=7.08\times10^{-10}\ C$

Distance = 4.0 cm

We need to calculate the electric field

Using for formula of electric field

$E=\dfrac{2q}{2\epsilon_{0}A}$

Where, q = charge

A = area

Put the value into the formula

$E=\dfrac{7.08\times10^{-10}}{8.85\times10^{-12}\times0.40}$

$E=200\ V/m$

We need to calculate the potential difference between the plates

Using formula of potential difference

$V=E\times d$

Where, E = electric field

d = distance

Put the value into the formula

$V=200\times0.04$

$V=8\ V$

Hence, The potential difference between the plates is 8 V.

3 0

4 years ago

OSCilloscope shows wave starts cycle minimum 20 units reaches max of 100 units before completing 5 milliseconds what is relation

Vitek1552 [10]

Answer:

$T=0.02\ s$

$f=50\ Hz$

Explanation:

Given:

minimum amplitude at the start of oscillation cycle, $a_0=20\ unit$

the first maximum amplitude after the start of oscillation cycle, $a_{m1}=100\ units$

Time taken to reach from the first minima to the first maxima, $t=5\times 10^{-3}\ s$

As we know that an oscilloscope executes a wave cycle represented by a sine wave. So we can deduce that it has executed one-fourth of the cycle in going from the amplitude of 20 units to 100 units in 0.005 seconds.

<u>So the time taken to complete one cycle of the oscillation:</u>

$T=4\times 0.005$