The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
Answer:
sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
Explanation:
We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis
So x component of the vector 
y component of the vector 
So vector will be 6.06i+3.5j
Now other vector of length of 7 units and makes an angle of 120° with positive x-axis
So x component of vector 
y component of the vector 
Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j
The time of motion of the track star is determined as 0.837 s.
<h3>Time of motion of the track star</h3>
The time of motion of the track star is calculated as follows;
T = (2u sinθ)/g
where;
- T is time of motion
- g is acceleration due to gravity
- θ is angle of projection
T = (2 x 12 x sin20)/9.8
T = 0.837 s
Learn more about time of motion here: brainly.com/question/2364404
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