Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
Answer:
an act of asking for information.
Answer:
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Answer: n the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7. Calculate the proper life of the ...
Explanation:
Answer:
F = 39.36 N
Explanation:
given,
initial speed, u = 38 m/s
final speed, v = 0 m/s
mass of ball = 0.145 Kg
time, t = 0.14 s
Force = ?
using impulse formula
J = change in momentum
J = F x t
m(v - u) = F x t
0.145 x (0 - (-38)) = F x 0.14
F x 0.14 = 5.51
F = 39.36 N
force exerted by the ball is equal to 39.36 N.