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solong [7]
3 years ago
7

A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r

educe his weight by 25% when he climbs. He wonders what the approximate size of such a balloon would be. Hearing of your physics skills, he asks you. Share with him your calculations that find the volume of the balloon.
Physics
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

V=16.65 m^3

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

R=\frac{25}{100}*74kg=18.5kg

So the net force uproad on the balloon is

F_b=18.5kg*g

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

P_h=\frac{m}{V}=0.179 kg/m^3

P_A=1.29 kg/m^3

F_b=F_A-F_H

F_b=m_a*g-m_h*g

m=P/V

18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g

V=\frac{18.5kg}{(1.29-0.179)kg/m^3}

V=16.65 m^3

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A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

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SI unit for electrical current
Mashcka [7]
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The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t
hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}    .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}    .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

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Answer:

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Explanation:

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Answer:

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