Speed of the tip of the minute hand=V=0.0244 cm/s
Explanation:
The angular velocity of the minute hand is given by
T= time period of the minute hand=60 min=3600 s
so ω= 2 π/3600 rad/s
Now linear velocity v= r ω
r= radius of minute hand=14 cm
so v= 14 (2 π/3600)
V=0.0244 cm/s
Answer:
a) 2.5 m/s²
b) 6.12 m/s
Explanation:
Tension of rope = T = 356N
Weight of material = W = 478 N
Distance from the ground = s = 7.5 m
Acceleration due to gravity = g = 9.81 m/s²
Mass of material = m = 478/9.81 = 48.72
Final velocity before the bundle hits the ground = v
Initial velocity = u = 0
Acceleration experienced by the material when being lowered = a
a) W-T = ma
⇒478-356 = 48.72×a
⇒a = 2.5 m/s²
∴ Acceleration achieved by the material is 2.5 m/s²
b) v²-u² = 2as
⇒v²-0 = 2×2.5×7.5
⇒v² = 37.5
⇒v = 6.12 m/s
∴ Velocity of the material before hitting the ground is 6.12 m/s
Answer:
a)
b)
Explanation:
Given data:
Electric field = 1.47 N/C
velocity of electron is
distance of point b from point A is 0.55 m
we know that acceleration of particle is given as
a) for electron
from equation of motion we have
b) for proton
from equation of motion we have