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Drupady [299]
3 years ago
15

Suppose you had used a less sensitive balance for the Archimedes method of getting volume of an object based on the difference i

n mass in water and out of water. Would that change the precisions of your density for that method?. They define sensitivity as the scale to which things are measured and precision as how repeatable something is.
Physics
2 answers:
alisha [4.7K]3 years ago
8 0
<span>If you think about it, changing the scale to which something is measured does not affect the repeatability of the measurement. For instance, if you have a meter stick which was labeled incorrectly, that doesn't affect the fact that every measurement you take of a certain fixed distance will still be the same. Precision does not equal accuracy.</span>
algol133 years ago
6 0

Answer:

There will be no change in the sensitivity of the method.

Explanation:

A less sensitive measuring method is less likely to record the small errors of measurement. The sensitivity of a measuring device depends on the error of measurement that it can accommodate. For example, is the length of a pen was to be measured by the ruler and a vernier caliper, the two measurements would be different. This is because the ruler would give a measurement to the nearest centimeter. However, the vernier caliper would provide a measurement to the nearest 0.1 cm which improves the accuracy and sensitivity of the measurement as well. Thus, the more sensitive an instrument is, the more accurate it is.

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A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the
Natali [406]
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2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

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4 0
3 years ago
A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm
iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

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m = 14*26 = 364

4 0
3 years ago
Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a
Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on R_3 \  and \ R_4, that is the same.

In option 2, this statement is false because Rcd=R_3+R_4 therefore it implies that Rcd is always larger then R_3.

In option 3,  this statement is true because the voltage of R_1 \ and \ R_2 is always equal.

In option 4, this statement is true because Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}is always smaller then 1 therefore,R_1 \times (\frac{R2}{(R1+R2)}) is always equal to R1.

5 0
2 years ago
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