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3 years ago

C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N

Physics

1 answer:

tiny-mole [99]3 years ago

8 0

Answer:

Spring cannot return to its original, since a part of its deformation is plastic, not elastic.

Explanation:

Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in elastic region, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.

Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is plastic, not elastic.

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An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$