Two forces FA and FB are applied to an object whose mass is 9.39 kg. The larger force is FA. When both forces point due east, the object's acceleration has a magnitude of 1.40 m/s2. However, when FA points due east and FB points due west, the acceleration is 0.950 m/s2, due east. Find (a) the magnitude of FA and (b) the magnitude of FB

Using Newton's second law: F=ma

m=9.39 kg

F=Fa+Fb=9.39*1.4=13.146
F=Fa-Fb=9.39*0.950=8.9205

Fa=(13.146+8.9205)/2=11.03 N
Fb=(13.146-8.9205)/2=2.11 N

Two forces and are applied to an object whose mass is 19.4 kg. The larger force is . When both forces point due east, the object's acceleration has a magnitude of 0.619 m/s2. However, when points due east and points due west, the acceleration is 0.501 m/s2, due east. Find (a) the magnitude of and (b) the magnitude of .

(a) Well, if you insist on pointing east, FA must contain all the cool moves. The magnitude of FA is 14.1 N, which is pretty awe-"sum".

(b) Now, FB wants to join the party but in reverse. To counterbalance FA, FB needs to be 9.51 N to make the acceleration zero. That's the power of symmetry, folks!

To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Let's denote the magnitude of FA as F and the magnitude of FB as F'.

In the first scenario, when both forces point due east, the object's acceleration is 1.40 m/s². Therefore, we have:

F - F' = m * a1

Substituting the given values, we have:

F - F' = (9.39 kg) * (1.40 m/s²)

Next, let's consider the second scenario, where FA points due east and FB points due west, and the acceleration is 0.950 m/s² due east. In this case, the forces are acting in opposite directions, so we have:

F + F' = m * a2

Substituting the given values again, we have:

F + F' = (9.39 kg) * (0.950 m/s²)

Now, we have a system of two equations:

F - F' = (9.39 kg) * (1.40 m/s²)
F + F' = (9.39 kg) * (0.950 m/s²)

We can solve this system to find the values of F and F'.

By adding the two equations together, we eliminate F' and obtain:

2F = (9.39 kg) * (1.40 m/s²) + (9.39 kg) * (0.950 m/s²)

Simplifying this equation, we have:

2F ≈ 26.226 + 8.901
2F ≈ 35.127

Dividing both sides of the equation by 2, we find:

F ≈ 35.127 / 2
F ≈ 17.5635

Therefore, the magnitude of FA, F, is approximately 17.5635 Newtons.

To find the magnitude of FB, F', we can substitute the value of F into one of the original equations. Let's use the equation F + F' = (9.39 kg) * (0.950 m/s²):

17.5635 + F' = (9.39 kg) * (0.950 m/s²)

Now, we can solve for F' by subtracting 17.5635 from both sides of the equation:

F' ≈ (9.39 kg) * (0.950 m/s²) - 17.5635

Calculating this, we find:

F' ≈ 8.901 - 17.5635
F' ≈ -8.6625

Since the magnitude of a force cannot be negative, we take the absolute value:

|F'| ≈ |-8.6625|
|F'| ≈ 8.6625

Therefore, the magnitude of FB, |F'|, is approximately 8.6625 Newtons.

To summarize:
(a) The magnitude of FA is approximately 17.5635 Newtons.
(b) The magnitude of FB is approximately 8.6625 Newtons.

A traveling wave has wavelength 0.50 m, speed 20 m/s. Find the wave frequency.