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bekas [8.4K]
3 years ago
8

Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

A - B, calculate the magnitude and angel with the x axis of the vector D
Physics
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

\vec a=(ax,ay)\ ,\  \vec b=(bx,by)

The sum of the vectors is:

\vec a+\vec b=(ax+bx,ay+by)

The difference between the vectors is:

\vec a-\vec b=(ax-bx,ay-by)

The magnitude of \vec a is:

|\vec a|=\sqrt{ax^2+ay^2}

The angle \vec a makes with the horizontal positive direction is:

\displaystyle \tan\theta=\frac{ay}{ax}\\

The question provides the vectors:

\vec a=(2,6)

\vec b=(2,22)

\vec d=\vec a-\vec b

Calculate:

\vec d=(2,6)-(2,22)=(0,-16)

The magnitude of \vec d is:

|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16

The angle is calculated by:

\displaystyle \tan\theta=\frac{-16}{0}

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁) is the mass of hockey player 1= 91.0-kg

(m₂) is the mass of hockey player 2=  91.0-kg

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(u₂) is the velocity before the collision of hockey player 2=?

a)

Momentum before the collision;

\rm  m_1u_1 + m_2u_2 \\\\ 91.0 \times 5.50 + 91.0 \times 8.1 \\\\ 1237.6 kg m/s^2

Momentum before the collision = 1237.6 kg m/s².

b)

The velocity of the two hockey players after the collision from the law of conservation of the momentum as:

Momentum before collision = Momentum after the collision

1237.6 kg m/s² = (m₁+m₂)V

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Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

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