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 (vmax) =?
Period of oscillation (T) =?
We know that
Fmax = mv2max/A
or , 0.8 = 200×10-3xv2max/20×10-3
or , 0.8×20/200= v2max
or , vmax = 0.28 m/s
Again,
vmax = ωA
or , v = 2π/T . A
or , T = 2πr/v =2πx20x10-3/0.28 =0.45s
∴ Maximum velocity = 0.28 m/s2
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/x2 ( air )
or , 0.1 =9×109xq2/(0.5)2
or , √0.1×0.25/9×109 = q
or ,q =√0.025/9×109
=√0.000278×10-8
=√2.78×10-12
∴ q =1.67×10-6 c
In the second part , a medium of dielectric constant is placed i.e. k =10 , Distance (r) =?
F =1/4π∈ok .q2/r2
or , 0.1 =9×109/10 x(1.67×10-6)2/r2
or , r =√9×109x(1.67×10-6)2
or , r =√25.10×10-3
∴r =0.16 m