Answer:
Explanation:
distance travelled
s = 2πR
= 2 X 3.14 X 140
= 880 m
final velocity = v
initial velocity = u
distance travelled = s
time = 60 s
s = ut + 1/2 at²
880 = .5 x a x 60²
a = .244 m/s²
final velocity v = at
= .244 x 60
= 14.66
centripetal acceleration at final moment
v² /R
14.66 X 14.66 / 140
= 1.53 m/s⁻²
1.53 m/s²
this is centripetal acceleration which acts towards the centre.
tangential acceleration calculated a _t = .244
redial acceleration ( centripetal ) = 1.53
Resultant acceleration
R²= 1.53² + .244 ²
R = 1.55 m/s²
total force = 1.55 x 76
= 118 N
<span>The correct answer is C) Two-second rule.
</span><span>
</span><span>This means that a driver must maintain a distance that would take at least 2 seconds to cover, at the current speed of the car. This means that in case the car in front suddenly comes to a full stop, the driver of the car behind has 2 seconds react and hit his own brakes. A two-car length rule might be alright when driving at low speeds, but when going at high speeds (on a highway, for example), a two-car length can be covered in a fraction of a second, which means a sudden braking by the car in front would almost guarantee an accident.</span>
The earth is the fifth largest planet in our solar system.
Answer: Magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
Explanation:
Given: Density = 80 
(1 n = 
m) = 
= 1.0 mm (1 mm = 0.001 m) = 0.001 m
= 3.0 mm = 0.003 m
r = 2.0 mm = 0.002 m (from the symmetry axis)
The charge per unit length of the cylinder is calculated as follows.
Substitute the values into above formula as follows.
![\lambda = \rho \pi (r^{2}_{2} - r^{2}_{1})\\= 80 \times 10^{-9} \times 3.14 \times [(0.003)^{2} - (0.001)^{2}]\\= 2.01 \times 10^{-12} C/m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Crho%20%5Cpi%20%28r%5E%7B2%7D_%7B2%7D%20-%20r%5E%7B2%7D_%7B1%7D%29%5C%5C%3D%2080%20%5Ctimes%2010%5E%7B-9%7D%20%5Ctimes%203.14%20%5Ctimes%20%5B%280.003%29%5E%7B2%7D%20-%20%280.001%29%5E%7B2%7D%5D%5C%5C%3D%202.01%20%5Ctimes%2010%5E%7B-12%7D%20C%2Fm)
Therefore, electric field at r = 0.002 m from the symmetry axis is calculated as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that magnitude of the electric field at a point which is 2.0 mm from the symmetry axis is 18.08 N/C.
