**Answer:**

**161.938 Hz**

**Explanation:**

the computation of the fundamental resonant frequency is shown below

p = 1100 kg/m^3

A = 130 mm^2

= 130 ×10^-6 m^2

T = 600 N

L = 20 cm

= 0.2 m

Now the linear density of tendon is

= 1100 kg/m^3 × 130 ×10^-6

= 0.143 kg/m

Now the wave of the string is

= √600 ÷ √0.143

= 64.775 m/s

Now finally the fundamental resonant frequency is

= 64.775 ÷ (2 × 0.2)

**=161.938 Hz**

Answer:

H h g hbgnbhvvhvhvnnnh hvh

Explanation:

J j hvdnhoighgfcbhvhvgfftj468 done ✔️

Answer:

Explanation:

Given that

Height of the tree is 3.7m

Therefore yo=3.7m

yox= 0m

The tiger lands 4.8m from the tree

Then, Range x=4.8m

Since she flies horizontally and she lands away from the bottom of the tree, her path will be trajectory as shown in the attachment

Let know the time of flight, using the equation of motion

voy is the initial velocity of the vertical motion of the is the tiger, which is zero at the beginning.

y = y0 + voy*t + ½*g*t²

Given that,

y=3.7m

yo=0m

voy=0m/s

g=9.81m/s²

y = y0 + voy*t + ½*g*t²

3.7=0+0•t+½×9.81×t²

3.7=0+0+4.905t²

3.7=4.905t²

t²=3.7/4.905

t²=0.7543

t=√0.7543

t=0.87sec

Time to reach the ground is

Now to know the initial velocity of the horizontal motion, using equation of motion

x=xo+Voxt

Vox is the horizontal initial velocity of the tiger.

x=xo+Voxt

4.8=0+Vox×0.87

4.8=0+0.87Vox.

4.8=0.87Vox.

Then, Vox=4.8/0.87

Vox=5.52m/s

Then, her initial velocity Vo

Vo=√voy²+vox²

Vo=√0²+5.52²

Vo=√5.52²

Vo=5.52m/s

The initial velocity is 5.52m/s

These sponsorships are somehow related to globalisation, as they are a global phenomenon, but some of them are on a very local scale, so globalisation is not a necessary part in them.

The best answer is c) commercialisation - atheletic competitions are turned into sources of profit, and athletes compete with the thought of winning money through it and the companies hope to increase their revenue by advertising during the sports events

From kinematic relation d = vi t + 1/2 at^2 , a = 2d/t^2 since it starts from rest . Acceleration does not depend on the mass. a = 2* 2.5 / (1.47)^2 = 2.3139 m/s^2.

Acceleration down the slide is due to the parallel component of gravity force. Friction opposes this force and acts up the inclined plane. mg sin theta - f = ma. Friction = f = (2.45 * 9.8) sin 30.5 - (2.45)(2.3139) = 6.5169 N.

Normal force = mg cos 30.5 = (2.45*9.8)* cos 30.5 = 20.6877 N. Friction = uk N. coefficient of friction = = friction / N = 6.5169 / 20.6877 = 0.32.

When it reaches the bottom of the slide, vf = vi + at = 0 + ( 2.3139)(1.47) = 3.4014 m/s = velocity at the bottom