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Anna11 [10]
3 years ago
11

A snail is at the bottom of a well, 115 feet deep. On day 1, it starts climbing up the side. That night it rests and slips down

one foot. It repeats this process, climbing up 3 feet each day and sliding down one foot each night until it reaches the top of the well. On what day number with the snail reach the top?
Physics
1 answer:
julia-pushkina [17]3 years ago
4 0

Answer:

Number of days needed = 57

Explanation:

Depth of well = 115 feet

Distance traveled in day = 3 feet up

Distance traveled in night = 1 feet down.

Total displacement in 1 day = 3 - 1 = 2 feet up

When the snail reaches 112 feet up, the next 3 feet up motion on day will make it reach the top.

\texttt{Days to travel 112 feet up = }\frac{112}{2}=56

After 56 days the displacement of snail is 112 feet.

On the next day it moves up by 3 feet and reaches out of well.

Number of days needed = 57

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Arada [10]

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3 years ago
I would like to know why this is the correct answer
Helen [10]

The acceleration of the object if the net force is decreased = 0.13 m/s²

<h3>Further explanation</h3>

Given

A net force of 0.8 N acting on a 1.5-kg mass.

The net force is decreased to 0.2 N

Required

The acceleration of the object if the net force is decreased

Solution

Newton's 2nd law :

\tt \sum F=m.a

The mass used in state 1 and 2 remains the same, at 1.5 kg

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ΣF=0.8 N

m=1.5 kg

The acceleration, a:

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