Answer:
taking a shower brushing your teeth and washing your hands
Explanation:
Since the new distance is 3 times the old distance,
the new force is (1/3²) = 1/9th of the old force.
That's kind-of Choice-D, but I really don't like the way choice-D is worded.
"9 times smaller" is really pretty meaningless.
Answer:
19.5°
Explanation:
The energy of the mass must be conserved. The energy is given by:
1) 
where m is the mass, v is the velocity and h is the hight of the mass.
Let the height at the lowest point of the be h=0, the energy of the mass will be:
2) 
The energy when the mass comes to a stop will be:
3) 
Setting equations 2 and 3 equal and solving for height h will give:
4) 
The angle ∅ of the string with the vertical with the mass at the highest point will be given by:
5) 
where l is the lenght of the string.
Combining equations 4 and 5 and solving for ∅:
6) 
Answer:
The magnitude of the net force is 5430N
Explanation:
I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

acting in the direction of the Drag, and the plane-perpendicular component is:

with negative sign since it counteracts the Lift.
So the components of the netforce F are:

The magnitude of the net force is:
