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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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Answer

a) For the rock

\dfrac{v_t^2sin 2\theta}{g} = \dfrac{v_t^2sin^2\theta}{2g}

2sin\thetacos\theta = \dfrac{sin^2\theta}{2}

2cos\theta = \dfrac{sin\theta}{2}

tan\theta = 4

\theta = tan^{-1} 4

\theta = 76^0

b) \theta = 45^0 for maximum range

\dfrac{d_{max}}{d}=\dfrac{(v_tcos 45^0)(2v_tsin 45^0)g}{(v_tcos 76^0)(2v_tsin 76^0)g}

\dfrac{d_{max}}{d}=\dfrac{0.707\times 0.707)}{0.97\times 0.242}

\dfrac{d_{max}}{d}=2.129

d_{max}=2.129 d

c) The value of θ is the same on every planet as g divides out.

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a roller coaster moves on a certain section of it's track with an average speed of 13 m/s. how much distance does it cover in 5.
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\\ \sf\longmapsto Speed=\dfrac{Distance}{Time}

\\ \sf\longmapsto Distance=Speed\times Time

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A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid
bixtya [17]

Answer:

u_{s}=0.56

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

F_{s.max}=\frac{mv^2}{r} \\

Where the r is the radius of merry-go-round and v is the tangential speed

but

F_{s.max}=u_{s}F_{N}=u_{s}mg

So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

Substitute the given values

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u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56

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