Find the ratio of the system r, and r₂ for m₁ = m₂ = 1 kg, k₁= 2000 N/m and k₂= 6000N/mBasek₁k₂00000m₁elleem₂O a. r₁=-31.88 andO b.₁ 1.1805 andO c.r₁=4.755 andO d. r₁= -0.0761 andx₁(1)r₂ = 20.91112 = -0.8471r2 = -61.34x₂ (1)12 = -5.541

To find: The ratio of the system. Given: The masses are m1=m2=1 kg. The spring constant are k1=2000…

Find the ratio of the system.r, and r₂ for m₁ = m₂ = 1 kg, k₁= 2000 N/m and k₂ = 6000 N/m Base k₁ k₂ 0xXxx00 m₁ 0x000 m₂ O a. r₁=-31.88 and O b. r₁=1.1805 and O c. r₁=4.755 and O d. r=-0.0761 and x₁(1) ₂= 20.911 2= -0.8471 2=-61.34 x₂(1) ₂= -5.541

Find the ratio of the system.r, and r₂ for m₁ = m₂ = 1 kg, k₁= 2000 N/m and k₂ = 6000 N/m Base k₁ k₂ 0xXxx00 m₁ 0x000 m₂ O a. r₁=-31.88 and O b. r₁=1.1805 and O c. r₁=4.755 and O d. r=-0.0761 and x₁(1) ₂= 20.911 2= -0.8471 2=-61.34 x₂(1) ₂= -5.541

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Properties of Fluids

The term fluid represents a type of substance that can easily change its shape under any external pressure/force and can flow. Fluids can acquire any shape in which it is stored and can resist normal stresses but cannot resist shear stress in rest condition. Whenever external shear stress is applied to fluids, it starts to flow. The fluids may be liquid or gas. Example- water, air, etc.

Question

Find the ratio of the system r, and r₂ for m₁ = m₂ = 1 kg, k₁= 2000 N/m and k₂= 6000
N/m
Base
k₁
k₂
00000
m₁
ellee
m₂
O a. r₁=-31.88 and
O b.
₁ 1.1805 and
O c.
r₁=4.755 and
O d. r₁= -0.0761 and
x₁(1)
r₂ = 20.911
12 = -0.8471
r2 = -61.34
x₂ (1)
12 = -5.541

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