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Flura [38]
4 years ago
5

He density of copper is 8.96g/cm^3 and the density of water is 1 g/cm^3. When a copper is submerged in a cylindrical beaker whos

e bottom has surface area 10 cm^2 the water level rises by 2 cm. The volume of the cylinder is the area of its base times its height. a) What is the specific gravity of copper?
b) What is the buoyant force on the copper object?
c) What is the buoyant force on the copper object?
d) What is the mass of the copper object?
Physics
1 answer:
Angelina_Jolie [31]4 years ago
6 0

Answer:

(a) 8.96

(b) 19600 dyne

(c) 19600 dyne

(d) 20 g

Explanation:

dcu = 8.96 g/cm^3, dw = 1 g/cm^3, A = 10 cm^2

Water level rises by 2 cm.

(a) The specific gravity of copper = density of copper / density of water

                                                       8.96 / 1 = 8.96

(b) According to the Archimedes's principle, the buoyant force acting on the body is equal to the weight of liquid displaced by the body.

Weight of water displaced by the copper = Area of beaker x rise in water level                          

                                                                      x density of water x gravity

                          = 10 x 2 x 1 x 980 = 19600 dyne

(c) Same as part b

(d) Let mass of copper be m.

For the equilibrium condition,

the true weight of copper = Buoyant force acting on copper

m x g =  19600

m = 19600 / 980 = 20 g

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vector u has a magnitude of 20 and direction of 0°.vector v has amagnitude of 40and a direction of 60°.find the magnitude and di
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Addition of vectors:

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a) Find the magnitude and direction of the resultant to the nearest whole

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The resultant of two vectors is simply the vector sum of the vectors. There are a handful of ways to present the resultant factor; the notation that shows the vector magnitude and direction is called the polar vector notation. An example of a vector presented in polar vector notation is

a∠θ where a is the magnitude and θis the angle that the vector makes with the horizontal axis.

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Let's first present the vectors in rectangular vector notation.

For the vector →u of magnitude 20 and direction 0∘ to the horizontal axis, the vector is →u=^i20.

For the vector →v

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The resultant vector →w is the vector sum of the vectors, i.e.

→w=→u+→v

=^i20+^i40cos60∘+^j40sin60∘

=^i(20+40cos60∘)+^j(40sin60∘)

=^i40+^j20√3

For a vector ^ix+^jy, the magnitude of the vector is √x2+y2 and the direction above the horizontal axis is θ=tan−1(yx).

Let's use the formulas:

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The magnitude of the vector is about 53 units in the direction 41-degrees above the horizontal axis.

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