Answer:
44.5s ; 22.64 m/s
Explanation:
The motion of the truck could be separated into 3 different phases :
First :
Time of motion :
Initial Velocity, u = 0 ; final velocity, v = 29 m/s
Acceleration, a = 2 m/s²
Recall: acceleration = change in velocity / time
Time = change in velocity / acceleration
Time = (29 - 0) / 2 = 14.5 second
Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m
Second :
Time = 25 seconds at constant speed
29 m/s for 25 seconds
v*t = 29 * 25 = 725 m
Third:
5 seconds before coming to rest
((29 + 0) /2) * 5
14.5 * 5 = 72.5 m
A.)
Length of journey = (14.5+ 25 + 5) = 44.5 seconds
B.)
Average velocity = total distance / total time taken
Average velocity = (210.25 + 725 + 72.5) / 44.5
= 1007.75 / 44.5
= 22.646067
= 22.64 m/s
Hey
in the summer it is because of the meting of snow
in the spring it is rain
hope l helped
<span>Target HR Zone 50-85% would be </span>90-153 beats per minuet. The average maximum heart rate 100% is 180 beats per minuet.
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Answer:
Explanation:
Length of bar = L
mass of bar = M
mass of each ball = m
Moment of inertia of the bar about its centre perpendicular to its plane is

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is


Total moment of inertia of the system about the centre of the bar perpendicular to its plane is
I = I1 + I2

