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Vector A has a magnitude of 5.00 units, and B has a magnitude of 9.00 units. The two vectors make an angle of \(50.0^{\circ}\) with each other. Find \(\mathbf{A} \cdot \mathbf{B}.\)

Short Answer

Expert verified
The dot product of vectors A and B is simply calculated using their given magnitudes and the cosine of the given angle, with the angle being converted to radians before calculation.

Step by step solution

01

Identify given quantities

First, identify the given quantities in the problem. The magnitude of vector A is 5.00 units, the magnitude of vector B is 9.00 units, and the angle between vector A and B is \(50.0^{\circ}.\)
02

Convert angle to radians

The angle given is in degrees but the cosine function in most calculators uses radians. So, convert \(50.0^{\circ}\) to radians. Using the formula, \(radians = degrees \times \frac{\pi}{180}\), we get \(50.0^{\circ} = 50\times \frac{\pi}{180} radians.\)
03

Compute the dot product

The formula for the dot product of two vectors is given by \(\mathbf{A}\cdot\mathbf{B} = |\mathbf{A}| \times |\mathbf{B}| \times \cos(\theta)\). So, substitute the known values into the formula to get \(\mathbf{A} \cdot \mathbf{B}= 5.00 \times 9.00\times \cos(50\times \frac{\pi}{180}).\)
04

Calculate the dot product

Finally, calculate the dot product using a calculator. Don't forget the degree to radians conversion when calculating the cosine.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle Conversion to Radians
When working with angles in mathematical equations, especially those using trigonometric functions like cosine, it's often necessary to convert angles from degrees to radians. This is because the standard units for angles in these functions are radians. Here's how you can do the conversion with ease.
Start with the formula:
  • Radians = Degrees × \( \frac{\pi}{180} \)
To convert 50 degrees to radians, multiply 50 by \( \frac{\pi}{180} \). This works because there are \( 2\pi \) radians in a full circle (360 degrees).
Thus,
  • \( 50^{\circ} \times \frac{\pi}{180} = 0.873 \) radians approximately.
Converting angles ensures your calculations are accurate when using a calculator, as these are typically set to handle radian measurements by default.
Vector Magnitude
Understanding the concept of vector magnitude is crucial when dealing with vector quantities. The magnitude of a vector is essentially its "length" or "size," regardless of its direction. For any vector \( \mathbf{V} \), the magnitude is denoted as \( |\mathbf{V}| \).
In our exercise, we have two vectors:
Vector \( \mathbf{A} \) with a magnitude of 5.00 units.
Vector \( \mathbf{B} \) with a magnitude of 9.00 units.
These numbers represent how long each vector is in the context of its application, such as force or velocity.
Remember these key points about vector magnitudes:
  • The magnitude of a vector can be calculated using the Pythagorean theorem for components, but in our problem, they are given directly.
  • The magnitudes are always non-negative.
  • In equations involving vectors, magnitude is crucial, as it scales the vector.
Understanding vector magnitude helps in calculating dot products, projecting vectors, and in various applications in physics and engineering.
Cosine Function
The cosine function is a key player when working with vector calculations, especially when determining the dot product. The cosine of an angle in a triangle is defined as the adjacent side divided by the hypotenuse. However, in vector math, it helps to find the angle between two vectors.
Here's how it applies in our context:
To compute the dot product, we use:
  • \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| \times |\mathbf{B}| \times \cos(\theta) \)
where \( \cos(\theta) \) is the cosine of the angle between the two vectors. This measures how aligned the vectors are; a cosine of 1 means they are perfectly aligned, whereas a cosine of 0 means they are perpendicular.
The cosine function can easily be computed using a calculator, remembering to input the angle in radians for the correct result.
Also note:
  • Cosine values range from -1 to 1.
  • It's periodic, with a period of \( 2\pi \).
Utilizing the cosine function effectively allows us to compute vector operations accurately and understand the spatial relationships between vectors.

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Most popular questions from this chapter

A 0.600 -kg particle has a speed of \(2.00 \mathrm{m} / \mathrm{s}\) at point (A) and kinetic energy of \(7.50 \mathrm{J}\) at point (B). What is (a) its kinetic energy at (A) ? (b) its speed at (B)? (c) the total work done on the particle as it moves from (A) to (B)?

A 100 -g bullet is fired from a rifle having a barrel \(0.600 \mathrm{m}\) long. Assuming the origin is placed where the bullet begins to move, the force (in newtons) exerted by the expanding gas on the bullet is \(15000+10000 x-25000 x^{2}\) where \(x\) is in meters. (a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel. (b) What If? If the barrel is 1.00 m long, how much work is done, and how does this value compare to the work calculated in (a)?

A bead at the bottom of a bowl is one example of an object in a stable equilibrium position. When a physical system is displaced by an amount \(x\) from stable equilibrium, a restoring force acts on it, tending to return the system to its equilibrium configuration. The magnitude of the restoring force can be a complicated function of \(x .\) For example, when an ion in a crystal is displaced from its lattice site, the restoring force may not be a simple function of \(x .\) In such cases we can generally imagine the function \(F(x)\) to be expressed as a power series in \(x,\) as \(F(x)=-\left(k_{1} x+k_{2} x^{2}+k_{3} x^{3}+\ldots\right) .\) The first term here is just Hooke's law, which describes the force exerted by a simple spring for small displacements. For small excursions from equilibrium we generally neglect the higher order terms, but in some cases it may be desirable to keep the second term as well. If we model the restoring force as \(F=-\left(k_{1} x+k_{2} x^{2}\right),\) how much work is done in displacing the system from \(x=0\) to \(x=x_{\max }\) by an applied force \(-F ?\)

A skier of mass \(70.0 \mathrm{kg}\) is pulled up a slope by a motordriven cable. (a) How much work is required to pull him a distance of \(60.0 \mathrm{m}\) up a \(30.0^{\circ}\) slope (assumed frictionless) at a constant speed of \(2.00 \mathrm{m} / \mathrm{s} ?\) (b) \(\mathrm{A}\) motor of what power is required to perform this task?

You can think of the work-kinetic energy theorem as a second theory of motion, parallel to Newton's laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a) and (b) separately from parts \((\mathrm{c})\) and \((\mathrm{d})\) to compare the predictions of the two theories. In a rifle barrel, a \(15.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(780 \mathrm{m} / \mathrm{s}\). (a) Find the work that is done on the bullet. (b) If the rifle barrel is \(72.0 \mathrm{cm}\) long, find the magnitude of the average total force that acted on it, as \(F=W /(\Delta r \cos \theta) .\) (c) Find the constant acceleration of a bullet that starts from rest and gains a speed of \(780 \mathrm{m} / \mathrm{s}\) over a distance of \(72.0 \mathrm{cm} .\) (d) If the bullet has mass \(15.0 \mathrm{g},\) find the total force that acted on it as \(\Sigma F=m a.\)

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