Answer:879.29 N-m
Explanation:
Given
mass of first child 
distance of first child from tree is 
tree is inclined at an angle of 
mass of second child 
distance of second child from tree is 
Weight of first child
Weight of second child
Torque of first child weight

Torque of second child weight

Net torque 
Answer:
d. None of the above.
Explanation:
In a parabolic motion, you have that in the complete trajectory the component velocity is constant and the vertical component changes in time. Then, the total velocity vector is not zero.
In the complete trajectory the gravitational acceleration is always present. Then, the grasshopper's acceleration vector is not zero.
At the top of the arc the grasshopper is not at equilibrium because the gravitational force is constantly acting on the grasshopper.
Then, the correct answer is:
d. None of the above.
100 J
Please mark me brainliest it would be greatly appreciated haha
Explanation:
Given that,
Initial speed of a car, u = 60 km/h = 16.67 m/s
Acceleration, a = 2m/s²
Final speed, v = 120 km/h = 33.33 m/s
We need to find the distance traveled and the time taken to make the distance.
acceleration = rate of change of velocity

let the distance be d.

Hence, the distance traveled and the time taken to make the distance is 208.25 m and 8.33 seconds respectively.
Answer:
TIME he applied the fertilizer to each plant
Explanation:
Independent variable in an experiment is the variable that is subject to change or manipulation by the experimenter. In this experiment, Bob wanted to investigate the effects of plant fertilizer. Bob sets up the experiment by applying the fertilizer to each plant at DIFFERENT TIMES i.e. plant 1-every morning, plant 2-once a week, plant 3-never.
Based on this, it is obvious that the independent or manipulated variable is the TIME at which he applied the fertilizer. On the other hand, the dependent or measured variable is the height of the plants.