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Lorico [155]
3 years ago
10

Starting from rest, a 88-kg firefighter slides down a fire pole. The average frictional force exerted on him by the pole has a m

agnitude of 800 N, and his speed at the bottom of the pole is 3.2 m/s. How far did he slide down the pole?
Physics
1 answer:
Cloud [144]3 years ago
3 0

Answer:h=7.22 m

Explanation:

Given

mass of firefighter m=88 kg

Frictional Force F=800 N

average speed at bottom v=3.2 m/s

Let h be the height of Pole

net force on Firefighter is

F_{net}=88\times 9.8-800=62.4

therefore net acceleration is

a_{net}=\frac{62.4}{88}=0.709 m/s^2

using v^2-u^2=2ah

here v=3.2 m/s

u=0

(3.2)^2-0=2\times 0.709\times h

h=\frac{10.24}{1.418}

h=7.22 m

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Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

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Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

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⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

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brainly.com/question/2903220

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