HELPPPPP MEEEEE PLEASE I NEED TO SUBMIT IN LESS THAN 10 MINSS
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Refer to the diagram shown below.
Let T = the tension in each wire.
For equilibrium,
2T cos(50°) = 150 N
1.2856T = 150
T = 116.677 N ≈ 117 N
Answer: 117 N
Let T = the tension in each wire.
For equilibrium,
2T cos(50°) = 150 N
1.2856T = 150
T = 116.677 N ≈ 117 N
Answer: 117 N
The kinematics of the uniform motion allows us to find the final position vector
r = (-41.575 i + 42.253 j) m
Given parameters
- the starting position x = -17.5 m y = 23.1 m
- jump time t = 10.7 s
- The average velocities vₓ = -2.25 m / s and v_y = 1.79 m / s
to find
- the final position
The uniform motion occurs when the velocity of the bodies is constant, in this case the relationship can be used for each axis
v =
x = x₀ + v t
Where vₓ it is the velocity, x the displacement, x₀ the initial position and t the time
Let's set a reference system with the horizontal x-axis. Regarding which we carry out the measurements
X axis
we look for the final position
x = x₀ + vₓ t
x = -17.5 -2.25 10.7
x = -41.575 m
Y Axis
we look for the final position
y = y₀ + v_y t
y = 23.1 + 1.79 10.7
y = 42.253 m
In conclusion, using the kinematics of uniform motion, find the final position vector
r = (-41.575 i + 42.253 j) m
learn more about uniform motion here: