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Archy [21]
3 years ago
8

Ari uses a tool called calipers to measure the diameter of different cylinders he is using in an experiment. His old calipers co

uld measure down to 0.01 mm, but his new calipers only measure down to 0.1 mm. Are his new calipers less precise, less accurate, both less accurate and less precise, or neither less precise or less accurate?
Physics
1 answer:
Vladimir [108]3 years ago
7 0

Answer:

Explanation:

Different measuring instruments have different level of precision . For example , our usual scale can measure upto 1 mm. Callipers can measure upto .1 mm. screw gauge can measure upto .01 mm.etc The deeper it can measure , the more precise it is. Hence screw gauge is most precise among the three.

On this basis we can say that , new caliper is less precise than old one.

Accuracy tells us about our result , how close our result is to the true measurement. It is dependent on  the manner in which  we take reading , So measuring methods , rather than measuring instruments , decide the accuracy of our result.

Hence we can say that the new caliper is less precise only.

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Answer:

The mass number of an element is the number of protons plus neutrons.

Explanation:

The mass number of an element is the number of protons plus neutrons, so the correct answer is the option C.

4 0
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Jack (mass 52.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. he collides with jill (mass 49.0 k
dybincka [34]
We must write down laws of conservation of momenta and energy. 
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Law of conservation of energy:
\frac{m_1v_1^2}{2}=\frac{m_1v'_1^2}{2}+ \frac{m_2v_2^2}{2}\\
m_2v_2^2=m_1v_1^2-m_1v_1'^2\\
v_2=\sqrt{\frac{m_1v_1^2-m_1v_1'^2}{m_2}
This will give us Jill's velocity after the colision.
v_2=6.43\frac{m}{s}
Law of conservation of momenta:
x: m_1v_1=m_1v_1'cos(34)+m_2v_2cos(\theta)\\
y: 0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\
We will use the second equation to get the angle at which the Jill is traveling:
0=m_1v_1'sin(34)-m_2v_2sin(\theta)\\
m_2v_2sin(\theta)=m_1v_1'sin(34)\\
sin(\theta)=\frac{m_1v_1'sin(34)}{m_2v_2}\\
\theta=sin^{-1}(\frac{m_1v_1'sin(34)}{m_2v_2})
When we plug all the number we get:
\theta=27.45^\circ
Please note that this is the angle below the x-axis.

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4 years ago
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Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
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