Answer:
left hand: 133 N upward
right hand: 83N downward
Explanation:
10 cm = 0.1m
Let g = 10m/s2. The pole gravity force would be 
. Suppose the pole mass distribution is uniform, this means the center of mass is at its geometric center, aka 5/2 = 2.5m from the top end.
In order for the system to stay balanced, the total net force and net moment must be 0
where 
are the forces exerted by his left and right hand, respectively.
We will pick his right hand (0.1 m from the top end) as moment pivot point. The total moment generated by pole gravity and his left hand must be 0
Distance (or moment arm length) from pole center to his right hand is 
Moment generated by gravity around his right hand is: 
counter-clockwise
Distance (or moment arm length) from left hand to his right hand is 
Moment generated by force from his left hand around his right hand is: 
clockwise
Since the total moment must be 0, the clockwised-direction moment = counterclockwised-direction moment:
upward
Therefore, the force exerted by his right hand must be 50 – 133 = -83 N or 83N downward
Answer:
66.053m/s
Explanation:
A = 47
B = 347
C = 19
Train moves at
(23 + A)m/s
= 23 + 47 = 60m/s
At (250.0+B) seconds
250.0+347 =
547 seconds
Distance d,
= 70 x 597
= 41790
It also moves at
(45.0 + c)
= 45 + 19
= 64m/s
Time = 800 + B
= 800 + 347
= 1147
Distance,
= 64 x 1147
= 73408m
Total distance,
= 73408 + 41790
= 115,198
Total time,
= 597 + 1147
= 1744
Average speed,
= Total distance / total time
= 115198/1174
= 66.053m/s
First solve the potential energy of the biker. using the fomula:
PE = mgh
where m is the mass of the object
g is the acceleration due to gravity ( 9.81 m/s2)
h is the height
PE = 96 kg ( 1120 m ) ( 9.81 m/s2)
PE = 1054771.2 J
then power = Work / time
P = 1054771.2 J / ( 120 min ) ( 60 s / 1 min)
P = 146.5 W
VF=0
Vi=60mph
T=4.0s
you need to convert hours to seconds... so you do there is 60 minutes in an hour and there is 60 seconds in a minute so you do 60*3=180
180/4.0=45m/s^2
