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egoroff_w [7]
3 years ago
7

A uniform solid cylinder of radius r and mass m can roll inside a hollow cylinder of radius R > r without slipping. A pendulu

m of length l = R−r 2 and mass M = m/2 is attached to the center of the smaller cylinder. Find the normal frequencies and normal modes of this system.

Physics
1 answer:
iogann1982 [59]3 years ago
8 0

Answer: The two answers are in explanation.

Explanation: Please find the attached files for the solution

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There is more land in the Northern Hemisphere than in the Southern Hemisphere. How might this difference affect CO2, concentrati
alexira [117]

Oxygen is a product of photosynthesis, therefor lover levels of C02 would be in the area with more water due to more phtosynthetic processes happening converting it into oxygen.

hope this helps, please mark brainliest. :)

5 0
3 years ago
The color of light most readily absorbed by water is _________.
zheka24 [161]

answer is the color white

3 0
3 years ago
Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.
Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is  1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(c) The energy lost when the coaster drops is 531,512.8 J.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

6 0
2 years ago
What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,
valina [46]
Both are constants used in the definition of Forces (gravitational and electric,respectively)

Since those constants are proportional to the magnitude of the forces:

Having a small gravitational constant explains why there is no apparent force of attraction with objects of considerable low mass (they would need to have great value of mass for the equation to give an apreciable force)

Electrical interactions are usually strong, and thus require an appropiate constant to depict the phenomenon. We deal in this case with charges really small, but the forces are in different order of magnitude.
5 0
3 years ago
Read 2 more answers
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