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Katena32 [7]
3 years ago
15

Set F1 = 5N at 0 degrees, F2= 5N at 90 degrees, F3 = 5N at 270 degrees and run the simulation. Using trigonometry, what net forc

e of F4 in the negative x-direction is necessary to produce an angle of 15 degrees between F2 and F3 and the y-axis? Set F4 to that value and run the simulation. Does the angle formed approximate 15 degrees?
Mathematics
1 answer:
Mrac [35]3 years ago
7 0

Answer: find the solution in the explanation

Step-by-step explanation:

Let's use resolution of forces by resolving into x - component and y- component.

X - component.

Sum of forces = F1 - F3 - F4cos 15

Sum of forces = 0

5 - 5 - 0.97F4 = 0

- 0.97 F4 = 0

F4 = 0

Y - component

Sum of forces = F2 + F4 sin 15

Sum of forces = 0

5 + 0.26F4 = 0

0.26 F4 = -5

F4 = -5/0.26

F4 = -19.23 N

Simulating F4 back into the equation

Sum of forces = F1 - F3 - F4cos 15

- F4cos Ø = 0

- (-19.23) cos Ø = 0

Cos Ø = 0

Ø = 1

Does the angle formed approximate 15 degrees ? NO

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M is a degree 3 polynomial with m ( 0 ) = 53.12 and zeros − 4 and 4 i . Find an equation for m with only real coefficients (i.E.
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Answer:

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

Step-by-step explanation:

Given that M is a polynomial of degree 3.

So, it has three zeros.

Let the polynomial be

M(x) =a(x-p)(x-q)(x-r)

The two zeros of the polynomial are -4 and 4i.

Since 4i is a complex number. Then the conjugate of 4i is also a zero of the polynomial i.e -4i.

Then,

M(x)= a{x-(-4)}(x-4i){x-(-4i)}

      =a(x+4)(x-4i)(x+4i)

      =a(x+4){x²-(4i)²}      [ applying the formula (a+b)(a-b)=a²-b²]

      =a(x+4)(x²-16i²)

      =a(x+4)(x²+16)      [∵i² = -1]

      =a(x³+4x²+16x+64)

Again given that M(0)= 53.12 . Putting x=0 in the polynomial

53.12 =a(0+4.0+16.0+64)

\Rightarrow a = \frac{53.12}{64}

      =0.83

Therefore the required polynomial is

M(x)=0.83(x³+4x²+16x+64)

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3 years ago
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