Given , Mass of the body (m) = 200g Amplitude (A) = 20mm Maximum force (F) = 0.8 N Maximum velocity (vmax) =? Period of oscillation (T) =? We know that Fmax = mv2max/A or , 0.8 = 200x10-3xv2max/20x10-3 or , 0.8x20/200= v2max or , vmax = 0.28 m/s Again, vmax = ωA or , v = 2π/T . A or , T = 2πr/v =2Read more

Given ,

Mass of the body (m) = 200g

Amplitude (A) = 20mm

Maximum force (F) = 0.8 N

Maximum velocity (v_{max}) =?

Period of oscillation (T) =?

We know that

F_{max }= mv^{2}_{max}/A

or , 0.8 = 200×10^{-3}xv^{2}_{max}/20×10^{-3}

or , 0.8×20/200= v^{2}_{max}

or , v_{max }= 0.28 m/s

Again,

v_{max }= ωA

or , v = 2π/T . A

or , T = 2πr/v =2πx20x10^{-3}/0.28 =0.45s

∴ Maximum velocity = 0.28 m/s^{2}

Time period =0.45s

Given, Force (f) =0.1 N Distance between Charge (r) =540 cm Charge (q) =? we know that , F = 1/4π∈o.qxq/x2 ( air ) or , 0.1 =9x109xq2/(0.5)2 or , √0.1x0.25/9x109 = q or ,q =√0.025/9x109 =√0.000278x10-8 =√2.78x10-12 ∴ q =1.67x10-6 c In the second part , a medium of dielectric constanRead more

Given,

See lessForce (f) =0.1 N

Distance between Charge (r) =540 cm

Charge (q) =?

we know that ,

F = 1/4π∈

_{o}.qxq/x^{2}( air )or , 0.1 =9×10

^{9}xq^{2}/(0.5)^{2}or , √0.1×0.25/9×10

^{9}= qor ,q =√0.025/9×10

^{9}=√0.000278×10

^{-8}=√2.78×10

^{-12}∴ q =1.67×10

^{-6 }cIn the second part , a medium of dielectric constant is placed i.e. k =10 , Distance (r) =?

F =1/4π∈

_{o}k .q^{2}/r^{2}or , 0.1 =9×10

^{9}/10 x(1.67×10^{-6})^{2}/r^{2}or , r =√9×10

^{9}x(1.67×10^{-6})^{2}or , r =√25.10×10

^{-3}∴r =0.16 m