Answer:
a. 6 seconds
b. 180 feet
Explanation:
Images attached to show working.
a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.
b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.
Answer:
Total time taken = 0.769 hour
Explanation:
using the velocity method
for sheet flow ;
Tt =
Tt = travel time
n = manning CaH
Pl = 25years
L = how length ( ft )
s = slope
For Location ( 1 )
s = 0.045
L = 1000 ft
n = 0.06 ( from manning's coefficient table )
Tt1 = 0.128 hour
For Location ( 2 )
s = 2.5 %
L= 750
n = 0.13
Tt2 = 0.239 hour
For Location ( 3 )
s = 1.5%
L = 500 ft
n = 0.15
Tt3 = 0.237 hour
For Location (4)
s = 0.5 %
L = 250 ft
n = 0.011
Tt4 = 0.165 hour
hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4
= 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour
Answer:
work done= 2.12 kJ
Explanation:
Given
N=2500 rpm
T=4.5 N.m
Period ,t= 30 s
P=
P=70,685W
P=70.685 KW
power=
work done = power * time
= 70.685*30=2120.55J
= 2.12 kJ