Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an angle of 55.0 ∘∘ above the negative x axis in the second quadrant. F⃗ 2F→2F_2_vec has a magnitude of 7.00 NN and is directed at an angle of 53.3 ∘∘ below the negative x axis in the third quadrant. Part A What is the x component of the resultant force? Express your answer in newtons.

-9.46 N

Explanation:

In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.

This value will be the projection of the force vector, on the x-axis.

For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:

cos θ =

In this case, x = F₁ₓ, and r = F₁

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º-55º = 125º

⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N

We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ =

In this case, x = F₂ₓ, and r = F₂

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º + 53.3º = 233.3º

⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N

The total component of both forces along the x axis, can be found just adding both components, as follows:

Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N

the answer to your question is high voltage and high currency

f=ma newton's second law 7x8=56n