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svetoff [14.1K]
3 years ago
13

Two cylindrical containers are placed on a turntable. canister A is empty; canister B contains lead shots. each canister is the

same distance r from the centre and the coefficient of static friction is uniform. what happens when the speed of the turntable is increased?
a) only the lighter container slides outward off the turntable; the heavier one stays on.
b) only the heavier container slides outward off the turntable; the lighter one stays on.
c) both containers slide off the turntable at the same turntable speed.
d) the lighter container slides inwards.
e) the heavier container slides inwards. ​
Physics
1 answer:
devlian [24]3 years ago
5 0

Answer:

c) both containers slide off the turntable at the same turntable speed.

Explanation:

The greater the mass of the cylinder, the greater the centrifugal force acting on it. That makes up for the greater force needed to move the heavier cylinder.

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Here are some examples of tropism in plants : Sunflowers are one of the best examples of positive phototropism , as they always grow facing the sun.

Explanation:

Although some experts consider that perhaps such a clear growth movement should not be considered tropism.

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A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i
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(a)  If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)  if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

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