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Sergeeva-Olga [200]
3 years ago
10

The first international standard of length was a metal bar kept at the International Bureau of Weights and Measures. One meter o

f length was defined to be the distance between two fine lines engraved near the ends of the bar. Why was it important that the bar be kept at a constant temperature?
Physics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

International Bureau of Weights and Measures have the standard bar for one meter length and it is made up of the platinium-iridium alloy. As these are metals so they have the thermal expansion. So the length of the bar depends upon the temperature. Temperature changes can effect the length of the standard bar. That's reason behind keeping the bar at the constant temperature as the change in temperature will result in the change of length.

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Selenium dichloride

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An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms
Ipatiy [6.2K]

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

v=u+at

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

v=0+10\times0.5

v=5\ m/s

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times10\times(0.5)^2

s=1.25\ m

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

v_{avg}=\dfrac{\Delta x}{\Delta t}

v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}

Where, x_{f}= final position

x_{i} = initial position

Put the value into the formula

v_{avg}=\dfrac{1.25+0}{0.5}

v_{avg}=2.5\ m/s

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7 0
3 years ago
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac
alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = 47\times 10^{-9}\ C/m^2

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We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

E=\dfrac{\sigma}{2\epsilon_0}

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}

Now, plug in  47\times 10^{-9}\ C/m^2 for 'σ' and 8.85\times 10^{-12}\ F/m for \epsilon_0 and solve for the electric field. This gives,

E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0
3 years ago
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