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finlep [7]
3 years ago
10

A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance

h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil.
a) How far up the smooth side will the marble go, measured vertically from the bottom?

b) How high would the marble go if both sides were as rough as the left side?

c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?
Physics
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

Explanation:

 a ) When the marble goes down , its potential energy is converted into both linear as well as rotational kinetic energy . So its total potential energy gets distributed between two components . Hence,  the component of linear kinetic energy which helps it to go up the right side wall gets reduced . As a result of it the height up to which it ascends on the right side smooth wall is comparatively less than h . At the top position on the right side smooth wall , it keeps on rotating only  having no linear upward motion there.

b ) When both sides are rough , marble will not  keep rotating on the right side at the top , because friction on the right wall  will stop it rotating . Hence it will be able to go upto the same height h as the left wall in this case.

c ) In case of absence of friction , it bears some of its energy in the form of rotational energy so height achieved by it becomes less. In case of friction , its rotational energy gets converted into linear kinetic energy , thus increasing the height achieved by it . In the whole process , it must be borne in mind that friction i not playing the role of dissipative force here because it is a case of perfect rolling and not slipping which causes dissipation.

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The speed of sound in aluminum is 5200 m/s. Can you hear a sound with a
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Answer:

v = wavelength * frequency

frequency = 5200 m/s / .2 m = 26000 / sec

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At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c
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Answer:

497.00977 N

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Explanation:

\rho = Density of water = 1000 kg/m³

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v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N

The drag force on the dolphin's nose is 497.00977 N

at 20°C

\mu = Dynamic viscosity = 1.002\times 10^{-3}\ Pas

Reynold's Number

Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005

The Reynolds number is 3742514.97005

8 0
3 years ago
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu
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Answer:

Acceleration, a=-2.48\ m/s^2

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It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}

a=-2.48\ m/s^2

So, the acceleration on the rough ice -2.48\ m/s^2 and negative sign shows deceleration.

8 0
3 years ago
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8090 [49]

Answer:

The new radius of the trajectory of the particle is four times the previous radius

Explanation:

In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:

r=\frac{mv}{qB}          (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

r'=\frac{mv'}{qB}=\frac{m(4v)}{qB}=4\frac{mv}{qB}=4r

The new radius of the trajectory of the particle is four times the previous radius

8 0
2 years ago
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