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Vitek1552 [10]
4 years ago
10

Consider the free-body diagram. Assume there is a net force of 30N to the right and the friction force equals 20N. What is the a

pplied force in this situation?
A.) 20N
B.) 30N
C.) 50N
D.) 80N

Physics
1 answer:
oksano4ka [1.4K]4 years ago
6 0

Answer:the answer would be C.50N

Explanation:I just took the test

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\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

We would calculate the magnitude by applying pythagorean theorem:

\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}

\longrightarrow \sf{Magnitude = 33.12}

\longrightarrow \sf{The \:  vector \: is \: (- 14, 30)}

The angle between two vectors is given by the formula:

\sf{\longrightarrow \small  \cos \emptyset =  \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }

In two dimensional, the x axis of vector form is:

\small\sf{\longrightarrow (b1, b2) = (1, 0) }

\sf{\longrightarrow \small \cos \:  \emptyset =  \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }

\small\longrightarrow \sf{ \dfrac{14}{33.12} }

\small\longrightarrow \sf{\emptyset  \:  = arcCos (\dfrac{ - 14}{33.12} )}

\small\longrightarrow \sf{\emptyset= 115^\circ}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\small\bm{The \: angle  \: between  \: the  \: vector \: }

\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}

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