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Yakvenalex [24]
3 years ago
7

The speed of a sound wave depends mostly on ?

Physics
1 answer:
jeyben [28]3 years ago
7 0
<span>The medium in which it travels through</span>
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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
3 years ago
What is transferred through sound waves?
alexandr402 [8]

Answer:

Sound waves transfer energy by causing successive compressions and rarefactions in the particles of the medium without transporting the medium particles themselves. Sound in solids can also manifest as transverse waves, causing crests and troughs in the propagation medium.

6 0
3 years ago
. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass
alisha [4.7K]

Answer:

10J

Explanation:

KE = (1/2)mv²

100J = (.5)(40kg)v²

v²=(100J)/(20kg)

v²= 5

KE = 5(.5)(4kg)

KE = 10J

3 0
2 years ago
Which direction does air circulate into low pressure zones in the northern and southern hemispheres
photoshop1234 [79]
Warm air goes up cool air goes down
5 0
3 years ago
Read 2 more answers
A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The
cluponka [151]

Answer:

v = 4.1 m/s

Explanation:

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

now for pure rolling condition we know that

v = R\omega

so we will have

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})

mgh = \frac{7}{10}mv^2

now we will have

v^2 = \sqrt{\frac{10}{7}gh}

v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}

v = 4.1 m/s

7 0
3 years ago
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