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kirill [66]
3 years ago
14

Three balls with the same radius 18 cm are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with

a density 842 kg/m^3 is held below the surface by a cord anchored to the bottom of the container, so that it is fully submerged. Ball 3, of density 1218 kg/m^3, is suspended from a rope so that it is fully submerged. Assume the density of water is 1000 kg/m^3 in this problem.
(a) Which is true for Ball 1?
MultipleChoice :
1) The magnitude of the buoyancy force is exactly one half of that of the ball's weight.
2) The magnitude of the buoyancy force is more than that of the ball's weight.
3) The magnitude of the buoyancy force is less than that of the ball's weight.
4) The magnitude of the buoyancy force is equal to that of the ball's weight.
(b) What is the tension on the rope holding the second ball, in newtons?
(c) What is the tension on the rope holding the third ball in N?
Physics
1 answer:
Kay [80]3 years ago
3 0

Answer:

(a)

4) The magnitude of buoyancy force is equal to that of ball's weight

(b) The magnitude of buoyancy force is larger than that of ball's weight. The tension on second ball is 158 newtons

(c) The magnitude of buoyancy force is larger than that of ball's weight. The tension on third ball is 218 newtons.

Explanation:

Newton's third law of motion states that forces always occurs in pairs. For every reaction there is an equal an opposite reaction. For Ball 1 the magnitude of buoyancy force is equal to that of ball's weight. Buoyancy force works against the gravity. Ball 2 and ball 3 have same buoyancy force. The buoyancy force for ball 2 and ball 3 is larger than that of ball's weight.

Tension = Wb - fb

Tension for Ball 2 = 1000 - 842 =  158 newtons

Tension for Ball 3 = 1000 - 1218 = -218 newtons

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A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
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Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

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velocity = 15m/s

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Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

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Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

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m2 = 2636.8 kg

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To solve this exercise we will use the concept related to heat loss which is mathematically given as

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Replacing

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