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3 years ago

An object in free fall is at heights y1, y2, and y3 at times t1, t2, and t3 respectively.

Physics

1 answer:

Dvinal [7]3 years ago

5 0

Answer:

a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

Explanation:

This is an exercise of average speed, which is defined with the variation of the distance in the unit of time

v = (y₃ - y₂) / (t₃-t₂)

the midpoint of a magnitude is the sum of the magnitude between 2

t_mid = (t₂ + t₃) / 2

the same reasoning is used for the mean acceleration

a = (v_f - v₀) / (t_f - t₀)

in our case

a = (v₃₂ - v₂₁) / (t₃₂ -t₂₁)

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Answer:

a).

Wmin= 2471.03 J

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$w= m*g\\W=m*g*h \\W=w*h$

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3 years ago

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

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2 years ago

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Answer:

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You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

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