Answer:
Accuracy
Explanation:
I think accuracy is more important. When it comes to vital organs in the body, the exactness of getting the measurement is paramount. Accuracy deals with getting very close, almost exact you may say, to a known standard. Precision on the other hand, deals with how easy a measurement can be retaken, reproduced or remade, irrespective of how far or close they are from the accepted norm.
From this, we can agree that precision neglects the most important factor, closeness or say, exactness. Precision isn't bothered by it. And while that can be excused in a few instances, it certainly can not be permitted when it comes to life, or organs of the body
Answer:
This question is incomplete
Explanation:
This question is incomplete because the telescope's focal length was not provided. The formula to be used here is
Magnification = telescope's focal length/eyepiece's focal length
The eyepiece's focal length was provided in the question as 0.38 m.
NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).
To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m
The potential difference is related to the electric field by:
∆V=Ed
where,
∆V is the potential difference
E is the electric field
d is the distance
what is potential difference?
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
We want to know the distance the detectors have to be placed in order to achieve an electric field of
E=1v/cm=100v/cm
when connected to a battery with potential difference
∆v=1.5v
Solving the equation,we find



learn more about potential difference from here: brainly.com/question/28166044
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Answer:
66w
Explanation:
p=w/t
p=660/10
p=66
prolly a bad explanation but hope it helps...
Answer:
1456 N
Explanation:
Given that
Frequency of the piano, f = 27.5 Hz
Entire length of the string, l = 2 m
Mass of the piano, m = 400 g
Length of the vibrating section of the string, L = 1.9 m
Tension needed, T = ?
The formula for the tension is represented as
T = 4mL²f²/ l, where
T = tension
m = mass
L = length of vibrating part
F = frequency
l = length of the whole part
If we substitute and apply the values we have Fri. The question, we would have
T = (4 * 0.4 * 1.9² * 27.5²) / 2
T = 4368.1 / 2
T = 1456 N
Thus, we could conclude that the tension needed to tune the string properly is 1456 N