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Lubov Fominskaja [6]
2 years ago
7

Two identical beakers with the same volume of water are placed on each pan of a double-pan balance. A steel ball is suspended fr

om a string and submerged in the water of one of the containers. A hollow plastic ball of the same volume is submerged in the water of the other container and fastened to the bottom of the beaker by a string. Will the balance move, and if so in which direction?
Physics
1 answer:
marusya05 [52]2 years ago
5 0

Answer:

The pan having beakers with steel ball will go down.

Explanation:

Both steel ball and plastic ball will experience buoyant force in upward direction . These forces will have same magnitude because they displace same volume of water. Both of them experience reaction force of buoyant force acting in downward direction which will try to lower the pan equally. But in case of plastic ball which is tied to the bottom , the tense  string will apply a force equal to buoyant force in upward direction on the bottom of the beaker.

This force will balance the reaction buoyant force acting on the bottom of beaker in downward direction . Hence this pan will remain balanced but the other pan will go down.  

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Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

L=2\pi R

where R is the radius of the wheel.

The period of revolution is:

T=120 s

Therefore, the tangential speed of the book is:

v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R

8 0
2 years ago
An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=584 kg, m2=838 kg, and m3=322 kg have blocke
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Answer:

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

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Total force required = Mass x Acceleration,

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Here we need to consider the system as combine, total mass need to be considered.

Total mass, a = m₁+m₂+m₃ = 584 + 838 + 322 = 1744 kg

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That is acceleration, a = 0.250 m/s²

Force required, F = ma = 1744 x 0.25 = 436 N

Force must be applied to m₁ to move the group of rocks from the road at 0.250 m/s² = 436 N

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Thus, 9130 joules of work is required to get the wheelbarrow across the yard.

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