Answer:
The acceleration of the car will be 
Explanation:
We have given that distance from stop sign s = 200 m
Time t = 0.2 sec
We have to find the constant acceleration
Now from second equation of motion 


So the acceleration of the car will be 
Answer:
Explanation:
The lift is going down with acceleration
Initial speed u = 0
Final speed v = 6 m/s
distance s = 15.25 m
acceleration a = ?
v² = u² + 2 a s
6² = 0 + 2 x a x 15.25
a = 1.18 m /s²
Elevator is going down with acceleration .
mg - T = ma where T is tension in the cable .
722 x 9.8 - T = 722 x 1.18
7075.6 - T = 851.96
T = 6223.64 N .
• 3 laps of 200 meter track = 600 meters
• 600 meters ran in 2 minutes
• 2 minutes = 120 seconds
• 600 meters / 120 seconds = 5 m/s on average
Answer:
42.69 N and 18.07 N
Explanation:
We are given that
Mass of ladder=6.2 kg
Length of ladder=1.97 m
Distance of Sawhorse A from one end=0.64 m
Distance of sawhorse B from other end=0.17 m
Let center of Ladder=
Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m
Distance of sawhorse B from center of ladder=0.985-0.17=0.815 m
Force one ladder due to gravity=mg=
Where 
Torque applied on Sawhorse A=
Torque applied on Sawhorse B=
In equilibrium


Total force=




