Can anyone help me with Combining like terms with negative coefficients I'm not good at math
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Answer:
Check attachment for necessary information
Step-by-step explanation:
At point B. Check attachment for free body diagram.
Where an is the centripetal acceleration and it is given as
an = Vb²/r
Fn = m•Vb²/r
Fn = 6 × Vb²/1.2
Fn = 5Vb²
Applying Newton's second law along the y direction
ΣF = m•ay
ay = 0, since the body is not moving in y direction
N —W = 0
N — WCosβ = 0
N = Fn = 5Vb²
5Vb²—58.86Cosβ = 0
Divide through by 5
Vb² — 11.772Cosβ = 0
Vb² = 11.772Cosβ, equation 1
Applying conservation of energy.
∆K.E(A) + ∆P.E(A) = ∆K.E(B) +∆P.E(B)
½m•Va²— 0+ mg•Ha — 0 = ½m•Vb² — 0 + mg•Hb — 0
½m•Va²+mg•Ha = ½m•Vb² + mg•Hb
From attachment
Ha = 1.13m
Hb = 1.2Cosβ.
Va = 2m/s²
½m•Va²+mg•Ha = ½m•Vb² + mg•Hb
½×6×2² + 6×9.81×1.13 = ½×6×Vb²+6×9.81×1.2Cosβ
12 + 66.512 = 3Vb² + 70.632Cosβ
From equation,.Vb² = 11.772Cosβ
78.512 = 3×11.772Cosβ+70.632Cosβ
78.512 = 35.316Cosβ+70.632Cosβ
78.512 = 105.948Cosβ
Cosβ = 78.512/105.948
Cosβ = 0.7410
β = ArcCos(0.7410)
β = 42.18 °
β = 42.2°
From the attachment, it is notice that,
θ + 20° = β
θ = β — 20°
θ = 42.2 — 20°
θ = 22.2°
The angle θ at which the box
left the smooth circular ramp is 22.2°
b. Using equation free fall motion
∆y = Vby•t + ½gt²
Let get Vb first, from equation 1
Vb² = 11.772Cosβ
Vb² = 11.772Cos42.2
Vb² = 8.721
Vb = √8.722
Vb = 2.95m/s
Now, to get Vby
Vby = VbSinβ
Vby = 2.95Sin42.2
Vby = 1.984 m/s
Then, applying free fall equation at point B
∆y = Vby•t + ½gt²
Hb - 0 = 1.984t + ½ × 9.81t²
1.2Cosβ = 1.984t + 4.905t²
1.2Cos42.2 = 1.984t + 4.905t²
4.905t² + 1.984t —0.889 = 0
Using formula method
t = [-b±√(b²-4ac)]/2a
a = 4.905 b = 1.984 and c = -0.889
t =[-1.984±√(1.984²-4×4.905×-0.889)] / 2×4.905
t = (-1.984±√21.378)/9.81
t = (-1.984± 4.624)/9.81
So,
t = (—1.984 — 4.624)/9.81
t = -0.674s
Or
t = (-1.984 + 4.624)/9.81
t = 2.64/9.81
t = 0.269s
Since time cannot negative, then,
t = 0.269s
Now, we can find the distance "s" by applying range formula, the part of motion is parabola this allow us to use projectile motion
R = Ux • t
s= Vbx × t
s= VbCosβ × t
s= 2.95Cos42.2 × 0.269
s= 0.588m
So, the distance "s" where the box fall into the cart is 0.588m
The correct answer for the problem shown in the figure atttached is the last option (option D), which is shown below:
D.(-x-9i)(x-9i)
If you want to verify that the option D.(-x-9i)(x-9i) is the correct answer, you can apply the distributive property to (-x-9i)(x-9i).
Therefore, when you apply the distributive property, you obtain that the result is:
-x²-81
The graph attached shown this equation, which is a parabola.
D.(-x-9i)(x-9i)
If you want to verify that the option D.(-x-9i)(x-9i) is the correct answer, you can apply the distributive property to (-x-9i)(x-9i).
Therefore, when you apply the distributive property, you obtain that the result is:
-x²-81
The graph attached shown this equation, which is a parabola.