a 20N mass is supported by two ropes. what is the tension in each rope? how woould i work this problem if i know the two angles
1 answer:
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1 kilometre=1000 metre
1 hour = 3600 second
The initial velocity of car A is 35.0 km/hr i.e
= 9.72 m/s
The initial velocity of car B is 45 km/hr =12.5 m/s
The initial velocity of car C is 32 km/hr = 8.89 m/s
The initial velocity of car D is 110 km/hr=30.56 m/s
The acceleration of car A is given as
The time taken by car A = 15 min.
From equation of kinematics we know that-
v= u+at [Here v is the final velocity and a is the acceleration and t is the time]
Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s
=11.456111106 m/s
The acceleration of B is given as
The time taken by car B =20 min
The final velocity of B is -
v= u+at
= u-at [Here a is negative due to deceleration]
=12.5 m/s +[0.0011574074074×20×60]
=13.8888888.....
=13.9
The acceleration of C is given as
The time taken by car C =30 min
The final velocity of C is-
v = u+at
=8.89 m/s+[0.003086419753×30×60] m/s
=14.4455555555..m/s
=14.45 m/s
The car C is decelerating.The deceleration is given as-
The time taken by car D= 45 min.
The final velocity of the car D is-
v =u+at
=30.56 -[0.00462962962962×45×60]m/s
=18.06 m/s
Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.
Considering conservation of mechanical energy we have:
K+U (start)= K+U (end)
Where:
K=kinetic energy=
U=potential energy=
So:
0.05*0.95*9.81+0 (start) = 0+K (end)
K (end)=0.5 Joules
K+U (start)= K+U (end)
Where:
K=kinetic energy=
U=potential energy=
So:
0.05*0.95*9.81+0 (start) = 0+K (end)
K (end)=0.5 Joules
Answer:
The cart moved away from the starting point between 8 s and 10 s.
Explanation:
Given that :
Which of these does NOT describe the cart’s motion on this graph?
The cart was at rest between 5 s and 7 s : From the distance of golf cart Vs time graph ; the car was at rest between 5s and 7s as the graph was flat with that time interval, meaning there was no change in distance.
The cart moved toward the starting point between 7 s and 12 s. : The graph depicts a negative slope at this time interval as the distance from starting point fell from about 26 m to 10 m
The cart moved away from the starting point between 8 s and 10 s. : At this time interval, the cart moved towards the starting not and not away. This could be seen in the decrease in Distance from starting point between the tune interval.
The cart moved away from the starting point between 2 s and 5 s. - - > The cart moved away from the starting point, with the positive slope signifying an increase in distance.
Therefore, The cart moved away from the starting point between 8 s and 10 s does not describe the motion of the cart on the graph.
