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3 years ago

3. A 92 kg Tarzan is holding on to a level 22m vine. He swings on the vine. What will his speed at the bottom of the swing be?

1 answer:

6 0

To solve this problem, we must always remember that energy is conserved. In this case, since he is falling down, he has highest potential energy at the top and zero at bottom. While his kinetic energy is zero at the top since he started from rest and highest at the bottom. We can also say that Potential Energy lost is Kinetic Energy gained thus,

- ΔPE = ΔKE ---> one is negative since PE is losing energy

- m g (h2 – h1) = 0.5 m (v2^2 – v1^2)

Where,

m = mass of tarzan (cancel that out)

g = gravitational acceleration

h2 = height at the bottom= 0

h1 = height at top = 22 m

v2 = velocity at the bottom

v1 = velocity at top = 0 (started from rest)

Therefore substituting all values:

- 9.8 (- 22) = 0.5 (v2^2)

v2 = 20.77 m / s (ANSWER)

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if length of the spring is doubled, what will happen to its time period? if mass of the spring is doubled and spring constant wi

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If the length of the spring is doubled, there will be no effect on its period.

If the mass of the spring is doubled and the spring constant is halved, then its period will be doubled.

The Simple Harmonic Motion is a type of periodic motion and one of its examples is the spring-mass system.

The period of a spring mass system is given by the following equation,

T= 2π√m/√k

Here, T= period of spring

m= Mass of the body attached to the spring

k= Spring constant

According to the above equation, the period of a spring depends on the mass of the body and the spring constant.

It is independent of the length of the spring.

Therefore, if the length of spring is doubled then there will be no effect on its period.

If the mass of spring is doubled and the spring constant is halved, then the equation becomes

T'= 2π√2m/√k/2

T'= √4 x 2π√m/√k

T'= 2 x T

Hence, if the mass of the spring is doubled and the spring constant is halved, its period will be doubled.

To know more about the "spring-mass system", refer to the following link:

brainly.com/question/13156044?referrer=searchResults

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1 year ago

A motorcycle is moving at a constant velocity of 15 meters/second. Then it starts to accelerate and reaches a velocity of 24 met

andrew11 [14]

Answer: $3 m/s^2$

Explanation:

The acceleration of the motorcycle is given by

$a=\frac{v-u}{t}$

where

v=24 m/s is the final velocity of the motorcycle

u=15 m/s is the initial velocity

t=3 s is the time taken

Substituting these numbers into the equation, we find

$a=\frac{24 m/s-15 m/s}{3 s}=\frac{9 m/s}{3s}=3 m/s^2$

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3 years ago

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Desiree has a scientific question that she wants to investigate. She has developed a hypothesis and method for her experiment. W

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The next scientific step is to collect data to test her Hypothesis.

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3 years ago

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Suppose a standing wave created on a spring that is 10.5 m long has a speed of 14 m/s and a frequency of 2 Hz. How many nodes an

dimulka [17.4K]

Answer:

If the two waves have the same amplitude and wavelength, then they alternate between ... In fact, the waves are in phase at any integer multiple of half of a period: ... The propagation velocity of the waves is 175 m/s.

Explanation:

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3 years ago

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

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3 years ago