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Pepsi [2]
3 years ago
11

A plaste horizontally at 150 m/s drops a package from an altitude of 1.3km. How far horizontally the package travel ?

Physics
1 answer:
tatuchka [14]3 years ago
6 0
How ya doing today sir I’m sure not too good cuz u ain’t gettin the answer sry
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A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m
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Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

H = \frac{u^{2}Sin^{2}\theta }{2g}

2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

K=\frac{1}{2}mv^{2}

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K = 13.5 J

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What does the "coefficient of friction" tell you?
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A bicycle wheel of radius 0.28 m rolls down a hill without slipping. Its linear velocity increases constantly from 0 to 8 m/s in
Marina86 [1]

Answer:

10.989 rad/s²

Explanation:

Angular Acceleration: This is the acceleration of a body that undergoes a circular motion. The S.I unit of angular acceleration is rad/s²

The expression for angular acceleration is given as

a = αr.............. Equation 1

Where a = linear acceleration, α = angular acceleration, r = radius.

make α the subject of the equation

α = a/r............. Equation 2.

But,

a = (v-u)/t ............. Equation 3

Where v = final velocity, u = initial velocity, t = time.

substitute equation 3 into equation 2

α = (v-u)/rt............ Equation 4

Given: v = 8 m/s, u = 0 m/s, r = 0.28 m, t = 2.6 s

Substitute into equation 4

α = (8-0)/(0.28×2.6)

α = 8/0.728

α = 10.989 rad/s²

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