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SOVA2 [1]
3 years ago
12

Could someone please explain to me Archimede's principle without using the term displace (or at least explaining what it means)?

Physics
1 answer:
disa [49]3 years ago
4 0
Does this help?

When an object is immersed in a fluid (in this case water, but may include both liquids and gases) the fluid exerts an upward force on the object which is called buoyancy force or <span>up-thrust.  Archimedes’ Principle states that the buoyant force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by that object.  Thus when an object is placed in water the rise in the water level is dictated by the mass of that object.</span>

<span>
</span>

<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>


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part (a) v\ =\ 10.42\ at\ 81.17^o towards north east direction.

part (b) s = 46.60 m

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  • velocity of the boat due to the north = v_b\ =\ 10.3\ m/s.

part (a)

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Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 90^o

Let 'v' be the velocity of the boat relative to the shore.

\therefore v\ =\ \sqrt{v_r^2\ +\ v_b^2}\\\Rightarrow v\ =\ \sqrt{1.60^2\ +\ 10.3^2}\\\Rightarrow v\ =\ 10.42\ m/s.

Let \theta be the angle of the velocity of the boat relative to the shore with the horizontal axis.

Direction of the velocity of the boat relative to the shore.\therefore Tan\theta\ =\ \dfrac{v_b}{v_r}\\\Rightarrow Tan\theta\ =\ \dfrac{10.3}{1.60}\\\Rightarrow \theta\ =\ Tan^{-1}\left (\dfrac{10.3}{1.60}\ \right )\\\Rightarrow \theta\ =\ 81.17^o

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total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken

Let 't' be the total time taken by the boat to cross the width of the river.\therefore w\ =\ v_bt\\\Rightarrow t\ =\ \dfrac{w}{v_b}\\\Rightarrow t\ =\ \dfrac{300}{10.3}\\\Rightarrow t\ =\ 29.12 s

Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river\therefore s\ =\ u_rt\\\Rightarrow s\ =\ 1.60\times 29.12\\\Rightarrow s\ =\ 46.60\ m

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