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Burka [1]
3 years ago
7

how do you find the uncertainty of a single measurement? For example, I am trying to find the uncertainty of 26 grams. ​

Physics
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

Please see answer below

Explanation:

The uncertainty will depend of the granularity of the instrument you use for the measurement.

For example, if you measure the length of a tabletop with a measuring tape with the smaller divisions given by the mm for example,then your measurement is limited by 1 mm (it cannot be better than that because your instrument doesn't have a finer division)

If you are using a digital scale to measure the weight of an object for example, and the scale displays in the readout 26.3 grams, then your uncertainty becomes 0.1 gram because that is the best your scale can read.

If on the other hand your scale reads no better than the gram, and gives you just 26 in the display, then the resolution of the scale is 1 gram, and that is what the uncertainty in your measurement is.

You have to report it as:  26 ± 1 g

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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
TEA [102]

Answer:

v=346.05\ m.s^{-1}

Explanation:

Given:

initial temperature of the lead bullet, T_i=43^{\circ}C

latent heat of fusion of lead, L_f=2.32\times 10^4\ J.kg^{-1}

melting point of lead, T_m=327.3^{\circ}C

We have:

specific heat capacity of lead, c=129\ J.kg^{-1}.K^{-1}

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)

\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f

\frac{1}{2} m.v^2=m(c.\Delta T+L_f)

\frac{v^2}{2} =129\times(327.3-43)+23200

v=346.05\ m.s^{-1}

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Which of the following accurately describes the behavior of these two mechanical waves when they intersect?
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A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t
levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

F=m\omega^2 r

where

m is the mass of the object

\omega is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s is the angular velocity

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Substituting, we find the force:

F=(0.040)(6.28)^2(0.30)=0.47 N

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3 years ago
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