Answer:
A) The resultant force is 43.4 [N]
B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis
Explanation:
We need to make a sketch of the different forces acting on the heavy crate.
In the attached image we can see the forces and the sum of the vector with their respective angles.
Forces in the X-axis

Forces in the y-axis
![FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]](https://tex.z-dn.net/?f=FDiony%3D0%5BN%5D%5C%5CFshirley%3D%2016.5%2Asin%2830%29%3D8.25%5BN%5D%5C%5CFjoany%3D19.5%2Asin%2860%29%3D16.88%20%5BN%5D%5C%5C%5C%5CForcesy%3D0%2B8.25-16.88%3D%20-8.63%5BN%5D)
Using the Pythagorean theorem

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.
An initiator is someone who starts something. Example: She initiated the fight.
Answer:
write the equation of motion go over the centre of mass
Explanation:
the center of mass of a distribution of mass in space (sometimes referred to as the balance point) is the unique point where the weighted relative position of the distributed mass sums to zero. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.
Answer:

Explanation:
As we know that initial speed of the fall of the stone is ZERO

also the acceleration due to gravity on Mars is g
so we have

now we have

now if the same is dropped for 4t seconds of time
then again we will use above equation



Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m