An empty jar is pushed open-side down into water so that trapped air cannot escape. As it is pushed deeper, the buoyant force on
1 answer:
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Answer:
Length of third side = 3.97m
Explanation:
First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.
From A-B we have distance = 5.5m = Say it c
From B-C we have distance = 4.3m = Say it a
From A-C we have distance = ? = Say it b which we have to find out.
Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)
For our case it is:
b² = a² + c² - 2acCos(B) - Say it equation 1
From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.
There values are:
a = 4.3m; c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees
Now by putting the respectve values in equation 1 we have:
b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)
b² = 18.49 + 30.25 - 32.95
b² = 15.79
b = √15.79
b = 3.97m
Thus the length of third side is 3.97m.
PS: The picture of triangle is being attached for yours understanding.
Answer:
a) V = 0.085 m^3
b) m = 17 kg
Explanation:
1) Data given
mb = 68 kg (mass for the block)
20% of the block volume is floating
100-20= 80% of the block volume is submerged
2) Notation
mb= mass of the block
Vw= volume submerged
mw = mass water displaced
V= total volume for the block
3) Forces involved (part a)
For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)
Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :
B = (mass water displaced) g = (mw) g (1)
The definition of density is :
If we solve for mw we got (2)
Replacing equation (2) into equation (1) we got:
(3)
On this case Vw represent the volume of water displaced = 0.8 V
If we replace the values into equation (3) we have
0.8 ρ_w V g = mg (4)
And solving for V we have
V = (mg)/(0.8 ρ_w g )
We cancel the g in the numerator and the denominator we got
V = (m)/(0.8 ρ_w)
V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3
4) Forces involved (part b)
For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.
We can begin finding the weight of the water displaced when the block is just about to sink (W1)
W1 = ρ_w V g
W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N
After this we can calculate the weight of water displaced before putting the bricks above (W2)
W2 = 0.8 x 833 N = 666.4 N
So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)
W3 = W1 -W2 = 833-666.4 N = 166.6 N
And finding the mass fro the definition of weight we have
m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg