Answer:
Answered
Explanation:
v= 1 m/s
A= 1 m^2
m= 100 kg
y= 1 mm
μ = ?
ζ= viscosity of SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2
forces acting on the block are
F_s ← ↓ →F_f
mg
N= mg
F_s= shear force = ζAv/y F_f= friction force = μN
now in x- direction F_s= F_f
ζAv/y = μN
0.3075×1×1×1/1×10^{-3} = μ×100
⇒μ=0.313 (coefficient of sliding friction for the block)
Now, as the velocity is increased shear force also increases and due to this frictional force also increases.
Now, to compensate this frictional force friction coefficient must increase
as v∝μ
d = 400 m (8 laps) = 2400 m
t = 14.5 min
v = 0
average t = 14.5 min (60 sec per min) = 870 s/8 laps = 108.75 s per lap
speed = distance /time
400/108.75 sec = 3.67 m/s
Answer:
t=0.704s
Explanation:
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?
from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed
Vy^2=0+2*9.8*1.73sin40.5
Vy^2=22.021
Vy=4.69m/s
Vx^2=u^2+2*9.81*cos40.5
Vy^2=25.81
Vy=5.08m/s
V=(Vy^2+Vx^2)^0.5
V=47.71^0.5
V=6.9m/s
from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.
f=force applied
v=velocity final
u=initial velocity
m=mass of the toy, 0.046
f=ma
f=m(v-u)/t
v=u+at
6.9=0+9.8t
t=6.9/9.81
t=0.704s