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3 years ago

8

Two wheels initially at rest roll the same distance without slipping down identical planes. Wheel B has twice the radius, but th

e same mass as wheel A. All the mass is concentrated in their rims so that the rotational inertias are I = mR2. Which has more translational kinetic energy when it gets to the bottom?

1 answer:

OlgaM077 [116]3 years ago

8 0

Answer:

Their translational kinetic energies are the same

Explanation:

The translational kinetic energy of an object is given by the formula:

$KE = 0.5 mv^2$

Where m = the mass of the object and

v = the linear speed of the object

From the question, it is stated that wheel A has the same mass as wheel B, that is $m_A = m_B$

Linear speed is also a function of the distance covered. Since both wheels cover the same distance within the same interval, we can conclude that $v_A = v_B$

Both wheels A and B have equal speed and mass, this means that their translational kinetic energy is the same.

Note that translational kinetic energy is not a function of the radius

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Answer:

<u>Very low</u>

<u>Explanation:</u>

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Bear in mind that for years some scientists have believed without any substantial evidence that there are other living beings in distant space.

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3 years ago

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

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Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

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substituting values

$P_w = 1000 * 9.8 * 0.3$

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While $P_o$ is the pressure of oil which is mathematically represented as

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substituting values

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$P_o = 882 \ Pa$

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3 years ago

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3 years ago

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denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

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a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

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t = time of observation

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b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

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<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

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$t_f=$ time of free fall

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$v_f=u+g.t_f$

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$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

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3 years ago

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Alla [95]

Answer:

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Explanation:

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3 years ago