Answer:
11.962337 × 10^-4 N
Explanation:
Given the following :
Length L = 11.8
Charge = 29nC = 29 × 10^-9 C
Linear charge density λ = 1.4 × 10^-7 C/m
Radius (r) = 2cm = 2/100 = 0.02 m
Using the relation:
E = 2kλ/r ; F =qE
F = 2kλq/L × ∫dr/r
F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))
2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4
In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214
Hence,
(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N
The best answer for this question is generally the best period to look for crabs is for the duration of the full moon, this is as a result of most invertebrates being nocturnal and being more effortlessly spotted and being fascinated to light sources.
Answer
given,
initial speed of the car = 90 m/s
final speed of the car = 0 m/s
distance taken to stop = 110 m
calculating the acceleration of the car = ?
using the equation of motion
v² - u² = 2 a s
0² - 90² = 2 x a x 100
a = 40.5 m/s²
acceleration of the formula one car is equal to a = 40.5 m/s²
For a pipe open at both ends, fund frequency (Fo) given byFo = v /2L
v = velocity of sound = 340 m/s (at about 20ºC)
Fo = 340/2*5.34 = 31.8 Hz
While for the one closed at one end and open at the other end...
For a pipe open at one end, fund frequency (Fo) given byFo = v/4L
Fo = 340/4*5.34= 15.9 Hz
