Q24P Vector A鈬赌which is directed alo... [FREE SOLUTION]

Chapter 3: Q24P (page 58)

Vector $\overset{鈬赌}{A}$ which is directed along an x axis, which to be added to
vector $\overset{鈬赌}{B}$ , which has a magnitude of 7.0 m.The sum is a third vector that
is directed along the y axis, with a magnitude that is 3.0 times that of $\overset{鈬赌}{A}$ . What is that magnitude of $\overset{鈬赌}{A}$ ?

Short Answer

Expert verified

Magnitude of $\overset{鈬赌}{A} = 2.2 m$

Step by step solution

To understand the concept

Using the vector algebra and simple mathematical operations, we can find the unknown vectors. herethe vector addition is to be used to find the magnitude of vector $\overset{鈬赌}{A}$ .

Vector $\overset{鈬赌}{A}$ can be written as

$\overset{鈬赌}{A} = x \hat{i} + y \hat{j}$ (1)

With magnitude

$|\overset{鈬赌}{A}| = \sqrt{x^{2} + y^{2}}$ (2)

Given

$|\overset{鈬赌}{B}| = 7.0 m (3) |\overset{鈬赌}{B} + \overset{鈫�}{A}| = 3 脳 |\overset{鈫�}{A}| (4)$

To find the magnitude of vector

Now,

$\overset{鈫�}{B} = B_{x} \hat{j} + B_{y} \hat{j} \overset{鈫�}{B} + \overset{鈫�}{A} = (B_{x} \hat{j} + B_{y} \hat{j}) + A \hat{i} \overset{鈫�}{B} + \overset{鈫�}{A} = (B_{x} + A) \hat{i} + B_{y} \hat{j}$

as the resultant is directed along y axis, so the x component will be zero.

$B_{x} + A = 0 B_{x} = - A B_{y} = 3 A \overset{鈫�}{B} = B_{x} \hat{i} + B_{y} \hat{j}$

Using equation (ii),

$7 = \sqrt{{(- A)}^{2} + {(3 A)}^{2}} 7 = A 脳 \sqrt{10} A = \frac{7}{\sqrt{10}} A = 2.2 m$

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Short Answer

Step by step solution

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To find the magnitude of vector

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91影视

Vector A鈬赌which is directed along an x axis, which to be added tovector B鈬赌, which has a magnitude of 7.0 m.The sum is a third vector that is directed along the y axis, with a magnitude that is 3.0 times that of A鈬赌. What is that magnitude of A鈬赌?

Short Answer

Step by step solution

To understand the concept

To find the magnitude of vector

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Force

Thermodynamics

Conservation of Energy and Momentum

Mechanics and Materials

Electricity and Magnetism

Solid State Physics

Study anywhere. Anytime. Across all devices.