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vladimir2022 [97]
3 years ago
6

Why centre of gravity of ring is out of its body

Physics
1 answer:
earnstyle [38]3 years ago
8 0
There's no reason why the center of gravity must be in a place where
there is a any mass. 

The center of gravity is simply a LOCATION ... the place where the 
amount of mass in any direction from it is the same amount. 

For that matter, whenever you know the location of the center of gravity
for ANY object, you can always go in there and scoop our a tiny spherical
hole at that place.  Then the center of gravity won't move, but it will be in an
empty space, 'outside the body' of the object.

Here are a few more points to ponder:

-- The center of gravity of a basketball, beach ball, tennis ball, or any other
inflated ball is the center of the ball, where there is no part of the skin.

-- The center of gravity of a party balloon is somewhere inside the balloon,
where there is no rubber.  If the balloon is spherical, then its center of gravity
is the center of the sphere.

-- The center of gravity of a square is the center of the square, not on any
of its sides.

-- The center of gravity of a triangle is the centroid of the triangle, not on any
of its sides.

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The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu
BartSMP [9]

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

8 0
3 years ago
When an astronomer sees certain stars and galaxies that look much redder than expected, what conclusion might the astronomer dra
Ratling [72]

It all comes to the doppler effect, the red shift means that the galaxy is moving away from us. The redshift is a result from the doppler effect, so as the galaxy moves away the wavelength expands, increasing the wavelength which responds to the red light.

8 0
3 years ago
Read 2 more answers
The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th
Scilla [17]

Length of the strings = 2.33 m

3 0
2 years ago
At the moment a hot cake is put in a cooler, the difference between the cakes and the cooler's
hram777 [196]

Answer: D(t)= 50(4/5)^t

Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).

If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.

5 0
3 years ago
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Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The long
nordsb [41]

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

R = \dfrac{\rho L }{A}

where;

A = πr²

R = \dfrac{\rho L }{\pi r ^2}

For the shorter cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{ 4 \rho L }{\pi \ D   ^2}

For the longer cylindrical resistor ; we have:

R = \dfrac{\rho L }{\pi r ^2}

since 2 r = D

R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}

R = \dfrac{32\rho L }{\pi \ (4 D)   ^2}

R = \dfrac{2\rho L }{\pi \ (D)   ^2}

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D   ^2}}{ \dfrac{2\rho L }{\pi \ (D)   ^2}}

\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D   ^2}}* { \dfrac  {\pi \ (D)   ^2} {2\rho L}}

\dfrac{R_s}{R_L} =2

{R_s}=2{R_L}

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

7 0
3 years ago
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