Answer:
v=14.14 m/s
t=1.141 s
Explanation:
Given that
h = 10 m
Initial velocity ,u = 0 m/s
We know that acceleration due to gravity g= 10 m/s²
Lets take final velocity = v
Final velocity v is given as
v² = u²+ 2 g h
v²= 0² + 2 x 10 x 10
v²= 200
v=14.14 m/s
Time taken t is given as
v= u + a t
a=acceleration
t=time
Now by putting the values in the above equation we get
14.14= 0 + 10 x t
14.14 = 10 t
t=1.141 s
Answer: (A) At terminal velocity ...
Explanation:
Answer:
298,220 N
Explanation:
Let the force on car three is T_23-T_34
Since net force= ma
from newton's second law we have
T_23-T_34 = ma
therefore,
T_23-T_34 = 37000×0.62
T_23= 22940+T_34
now, we need to calculate
T_34
Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2
F= ma
So, F= 12×37000×0.62= 22940×12= 275280 N
T_23 =22940+T_34= 22940+ 275280= 298,220 N
therefore, the tension in the coupling between the second and third cars
= 298,220 N
Answer:
8
Explanation:
Applying,
v = λf................ Equation 1
Where v = velocity/ speed of the wave, λ = wave length of the wave, f = number of waves the person surf in one seconds.
make f the subject of the equation
f = v/λ............ Equation 2
From the question,
Given: v = 1.6 m/s, λ = 24 m
Substitute these values into equation 2
f = 1.6/24
f = 0.0667 wave/seconds.
If,
in one seconds, the person surf a total wave crest of 0.0667
Therefore,
in one hours, he will surf a total wave crest of (0.0667×60×60) = 240 waves crest
He rides for every 30th wave crest,
Hence,
number of wave crest the person surf in one hour = 240/30 = 8
