Grade 12Mechanics

A meson is shot with constant speed 5.0 × 10 6 m/s in an electric field which produces on the meson an acceleration of 1.25 × 10 14 m/s 2 directed opposite to the initial velocity. How far does the meson travel before coming to the rest ?

Tamil selvi

8 Years agoGrade 12

3 Answers

Khimraj

8 Years ago

The distance covered by the meson in the electric field is given by 3^rd equation of motion.

v² = u²+2as

0 = 25*10¹² -2*1.25*10¹⁴s

So s = 1/10m = 10cm

Rashmi

4 Years ago

Given,

Initial velocity (u)=5.0×10⁶m/s

Acceleration (a) =1.25×10⁴m/s^2

Final velocity (v) = 0

Then,

According to the third equation of motion,

^{V^2=u^2+2as}

^{0=(5×10^6)}^{^2 +2×1.25×10^4×s}

^s=1/10m

^Or

^{S=10 cm}

Sudhin Varghese

4 Years ago

Given that

The acceleration of meson = 1.25 × 10 14 m/s²

The initial velocity (u) = 5.0×106m/s

The final velocity (v) = 0

We know that the third equation of motion states that,

V²=u²+2as

Putting values,

0=(5×10⁶)² + (2×1.25×10⁴×s)

0 = (25×10¹²) + (2.5×10¹⁴× s)

s=1/10 m or 10 cm

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View all Questions

A meson is shot with constant speed 5.0 × 10 6 m/s in an electric field which produces on the meson an acceleration of 1.25 × 10 14 m/s 2 directed opposite to the initial velocity. How far does the meson travel before coming to the rest ?

Tamil selvi

3 Answers

Khimraj

Rashmi

Sudhin Varghese

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