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erik [133]
3 years ago
9

While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?

Physics
1 answer:
mote1985 [20]3 years ago
5 0

While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

<span>(a)    </span>The speed of your phone after 4 s is:

V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

<span>(b)   </span>The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

<span>(c)    </span>If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

<span> </span>

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alexandr402 [8]

Answer:

the amplitude of  the subsequent oscillations is 0.11  m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, m_{1} = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

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the speed of glider,  v_{2} = 170 cm/s = 1.7 m/s

the amplitude of  the subsequent oscillations is A = 0.11  m

according to mechanical enery equation, we have

A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}

where

A is the amplitude and  v_{f} is the final speed.

to find v_{f}, we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

P_{f} = P_{i}

(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}

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0.95 v_{f} = 0.34

v_{f} = 0.36 m/s

Now we can calculate the amplitude

A = \sqrt{\frac{0.75 +0.2 }{10} }0.36

A = 0.11  m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π\sqrt{\frac{m_{1}+m_{2}  }{k} }

T = 2π\sqrt{\frac{0.75+0.2  }{10} }

T = 1.94 s

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