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ale4655 [162]
2 years ago
12

A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the

magnitude of A?
a)11
b)7.1
c)5.1
d)8.3
e)12.2
Physics
2 answers:
snow_tiger [21]2 years ago
5 0

The correct answer would be letter d, 8.3.

 

Solution for the problem follows:

 

Given are:

B = 6i - 8j 


A is unknown; let A be = mi + nj 



A+B is along the x axis (therefore A+B = Ki + 0j, where K is unknown,

 

but then again the magnitude of A+B is the similar as the magnitude of A,

 

so mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 



A+B, from simple vector addition, will be now (m+6)i + (n-8)j.

 

Ever since we previously know A+B = Ki + 0j, we now know that: 

m+6 = K 


n-8 = 0, which implies n=8. 

Thus, K^2=m^2+n^2 ====> (m+6)^2 = m^2 +8^2 


= m^2 + 12m + 36 = m^2 + 64 


= 12m = 28 


= m = 2.33333... 

Therefore, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333

Eva8 [605]2 years ago
5 0

<u>The magnitude of \vec A is 8.3.</u>

Further explanation:

A vector is a quantity having magnitude and direction both. It is represented as the product of magnitude and direction vector.

Given:

The vector is given as \vec B = 6\hat i - 8\hat j.

The direction of resultant vector is in the X- direction.

Concept used:

Consider a vector \vec A = a\hat i + b\hat j which is added to \vec B to get a resultant vector \vec C. The resultant vector is directed in the positive X-direction which means that the y- component of the vector is zero.

The expression for the resultant vector is given as.

\vec C = \vec A + \vec B

 

Substitute 6\hat i - 8\hat j for \vec B in the above expression.

\begin{gathered}\vec C = \left( {a\hat i + b\hat j} \right) + \left( {6\hat i - 8\hat j} \right) \\= \left( {a + 6} \right)\hat i + \left( {b - 8} \right)\hat j \\ \end{gathered}

 

The resultant vector is represented as.

\vec C = x\hat i + 0\hat j

 

Compare the above two expression of resultant vector.

x = a + 6                                         …… (1)

\begin{aligned}0&=b-8\\b&=8\\\end{aligned} 

The magnitude of resultant vector is equal to the magnitude of \vec A.

The expression for the magnitude of \vec A is given as.

\left| {\vec A} \right| = \sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}                           …… (2)

The expression for the magnitude of \vec C is given as.

\left| {\vec C} \right| = \sqrt {{{\left( {a + 6} \right)}^2}}

 

Compare the above two expressions.

\sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}= \sqrt {{{\left( {a + 6} \right)}^2}}

 

Substitute 8 for b  in above expression.

\begin{aligned}\sqrt {\left( {{a^2}} \right) + \left( {{8^2}} \right)}&=\sqrt {{{\left( {a + 6} \right)}^2}}\\{a^2} + 64&={a^2} + 36 + 12a \\12a&=28 \\a&=2.33 \\ \end{aligned}

Substitute 2.33 for a and 8 for b in equation (2).

\begin{aligned}\left| {\vec A} \right|&=\sqrt {{{\left( {2.33} \right)}^2} + {{\left( 8 \right)}^2}}\\&=8.33 \\ \end{aligned}

 

Thus, the magnitude of vector A is 8.33.

Learn more:

1.  Motion under friction brainly.com/question/7031524.

2.  Conservation of momentum brainly.com/question/9484203.

3. Circular motion brainly.com/question/9575487.

Answer Details:

Grade: College

Subject: Physics

Chapter: Vectors

Keywords:

Vectors, product, magnitude, direction, resultant vector, adding vector, subtraction of vector, 2.33, 8, 8.33, 8.3, 8.33.

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We will solve this problem by equating forces on each axis.

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While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

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⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

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⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

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