Your answer for part a is correct. Using Newton's 3rd Law, the force on the rocket by the exhaust leaving the rocket is the same magnitude (opposite direction) as the downward force applied to the exhaust.
In part b the net force is the upward force from the exhaust thrust minus the downward force of gravity:
Then using the Second Law, we get for part c:
The force and acceleration are in the upward direction
Answer:
A force of 75 N placed at 0.7 m on the meter stick.
Explanation:
The weight of the box is equal to:
The net torque is equal to zero and is equal to:
For a force with a value of 75 N that is placed at 0.70 m on the meter stick, it would produce a torque of 15 N m
If you replace that values in the equation:
India's monsoon area i beleive
Answer:
709.06 N
Explanation:
To do this, you need to do the drawing of this exercise. In the picture below, you have the draw, and the procedure to do this exercise. You need to apply 2nd Newton's law which is:
F = m*a
However, in this case, we only have tension and the weight of the object (As you can see in the picture), and we have no movement, the object it's supposed to be in rest, so, second law becomes this:
F = 0
In this case, we have rod and weight so the forces involving here are those, so second law is like this:
T1y + T2y - W = 0
To get the T1y and T2y, we have the angle so:
T1y = 320 sin80° = 315.14 N
T2y = 400 sin80° = 393.92 N
With these values, replace them and we get:
315.14 + 392.92 - W = 0
709.06 - W = 0
W = 709.06 N
Part 1)
here we know that supply took 10 s to reach the ground
so here we will have
Part 2)
Here all the supply covered horizontal distance of 650 m in 10 s interval of time
so here we can say
