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Phantasy [73]
3 years ago
13

One of the most effective ways to evaluate data is to try to replicate it.

Physics
2 answers:
WINSTONCH [101]3 years ago
6 0
It seems that you have missed the given options for the given statement above whether it is true or false. But anyway, the correct answer would be TRUE. It is true that one <span>of the most effective ways to evaluate data is to try to replicate it. Hope that this answer will help you. </span>
Dominik [7]3 years ago
3 0

Answer:

The correct answer will be-true

Explanation:

The evaluation of any data relies on two key components: the reproducibility and the replicability of the data which help understand the phenomenon.

The replicability of the data increases the chances of the data as it provides the sample size which can support the existing data also reduces the error along with improving the efficiency and accuracy of the experiment.

Therefore replication of the data is important to any scientific process and thus, true is the correct answer.

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An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r
alexandr402 [8]
Hi there, 
for this question we have:
Signal 2.0 MHz = Emitted so we can call it f_e
and we need the Reflected = f_{r}
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted). 
we also know: ΔBeat frequency(max) = 560 Hz = f_{b}
so we have: 
f_{e} - f_{r} = f_{b}
so frequency of Reflected is: 
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = f_{r}
now you know that Lambda = v/f 
so if we find the lambda with our Emitted then we can find v with the Reflected: 
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m 
=> v_{max} = (lambda)(f_{r} 
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s 
so the v_{max} is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0
4 years ago
Three different people weight a standard mass of 2.00 g on the same balance. Each person obtains a reading of 2.32 g for the mas
Lubov Fominskaja [6]

Answer: precise

Explanation:

Three different people weight a standard mass of 2.00 g on the same balance. Each person obtains a reading of 2.32 g for the mass of the standard. These results imply that the balance that was used is precise.

Precision can be defined as the closeness of measured values to each other, for a measuring equipment it is the closeness of the values of readings obtained at different times to each other. It does not necessarily means the measurements are accurate(closeness to the actual value). Therefore, in the case above where three different people measured the same mass on the same balance, and each of them obtained the same value which is different from the standard value. We can say the balance used is precise because the three readings are the same.

5 0
3 years ago
What is a characteristic of weight &lt;br /&gt;<br>I need it please answer it​
Marta_Voda [28]

Answer:

Weight is a consequence of the universal law of gravitation

Explanation:

Weight, gravitational force of attraction on an object, caused by the presence of a massive second object, such as the Earth or Moon. Weight is a consequence of the universal law of gravitation: any two objects, because of their masses, attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

8 0
3 years ago
A dumped harmonic oscillator of a mass of 500 g has a period of 0.5 second. The amplitude of the oscillation is decreasing 2.0 %
inn [45]

Answer:

The answer is below

Explanation:

The amplitude decreases by 2%  during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:

y = abˣ; where x ad y are variables, a is the initial value and b is the factor.

Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:

a = 10 cm, b = 2% = 0.02.

Therefore:

y = 10(0.02)ˣ

The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:

y = 10(0.02)²⁵

y= 3.355 * 10⁻⁴² cm

The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm

7 0
3 years ago
When the volume increased, what happened to the density?
dlinn [17]

the density decreases

7 0
4 years ago
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