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Vera_Pavlovna [14]
2 years ago
8

A beaker is filled to the brim with water. A solid object of mass 3.00 kg is lowered into the beaker so the object is fully subm

erged in the water (see the drawing). During this process, 2.00 kg of water flows over the rim and out of the beaker. What is the buoyant force that acts on the submerged object, and, when released, does the object rise, sink, or remain in place?
Physics
1 answer:
Sholpan [36]2 years ago
6 0

Answer:

Bouyant Force is 19.6N and the object will sink

Explanation:

The buoyant force is equal to the weight of displaced water(Archimedes Principle). Since 2.00 kg of water were displaced, the buoyant force would be equal to:

Fb = Fw = mg = (2.00 kg)(9.81 m/s²) = 19.6 N

Where Fb is buoyant force

Where Fw is weight of displaced water

The object will float if the buoyant force is greater than its weight, and sink if the reverse is true. Since the object's mass is 3.00 kg, its weight is:

Fw = mg = (3.00 kg)(9.81 m/s²) = 29.4 N

Since the object's weight, 29.4 N, is greater than the weight of displaced water, the object will sink.

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-10.9 rad/s²

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mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

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Where Xf is the final position of the cart and Xi the initial position of the cart.

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The cart comes to a stop before changing direction

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Answer:

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Explanation:

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The initial velocity is

y'(0) = 0

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