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alexandr402 [8]
3 years ago
6

You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You p

lace the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that
a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).  

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If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?
Ganezh [65]

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5 0
3 years ago
A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe
pentagon [3]

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0
3 years ago
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