Evaluate the cross product which \(\left( {5\widehat I + 3\widehat J} \right) \times \left( { - 4\widehat I + 2\widehat J} \right),\)expands to\( - 20\widehat I \times \widehat I + 10\widehat I \times \widehat J - 12\widehat J \times \widehat I + 6\widehat J \times \widehat J\).

Short Answer

Expert verified

The cross product is -\( - 20\left( {\widehat I \times \widehat I} \right) + 10\left( {5\widehat I \times 2\widehat J} \right) + 3\widehat J \times \left( { - 4\widehat I} \right) + \left( {3\widehat J \times 2\widehat J} \right)\)

Step by step solution

01

Given data

Given is the cross product\(\left( {5\widehat I + 3\widehat J} \right) \times \left( { - 4\widehat I + 2\widehat J} \right)\)

02

Definition of cross product

The cross product is a binary operation on two vectors in three-dimensional space. It creates a vector that is perpendicular to both vectors. The vector product of two vectors, a and b, is represented by a, b. It generates a perpendicular vector to both a and b. Vector items are also known as cross goods. The cross product of two vectors produces a vector that may be calculated using the Right-hand Rule.

If \(\overrightarrow A \)and \(\overrightarrow B \) lie in the \(xy\) plane, we can use the results for the unit vectors to calculate the cross product, which will be in the \( + z\)or \( - z\) direction

\(\begin{aligned}{}\overrightarrow A \times \overrightarrow B &= ({A_x}\widehat i + {A_y}\widehat j) \times ({B_x}\widehat i + {B_y}\widehat j)\\ &= \left( {{A_x}{B_x}} \right)\widehat i \times \widehat j + \left( {{A_y}{B_y}} \right)\widehat j \times \widehat j + \left( {{A_x}{B_y}} \right)\widehat i \times \widehat j + \left( {{A_y}{B_x}} \right)\widehat j \times \widehat i\end{aligned}\)

03

Evaluate the given cross product

The given cross product can be evaluated as below:

\(\begin{aligned}{}\left( {5\widehat I + 3\widehat J} \right) \times \left( { - 4\widehat I + 2\widehat J} \right) &= 5\widehat I \times \left( { - 4\widehat I + 2\widehat J} \right) + 3\widehat J \times \left( { - 4\widehat I + 2\widehat J} \right)\\& = \left( {5\widehat I \times \left( { - 4\widehat I} \right)} \right) + \left( {5\widehat I \times 2\widehat J} \right) + 3\widehat J \times \left( { - 4\widehat I} \right) + \left( {3\widehat J \times 2\widehat J} \right)\\ &= - 20\left( {\widehat I \times \widehat I} \right) + 10\left( {5\widehat I \times 2\widehat J} \right) + 3\widehat J \times \left( { - 4\widehat I} \right) + \left( {3\widehat J \times 2\widehat J} \right)\end{aligned}\)

Hence, the product is \( - 20\left( {\widehat I \times \widehat I} \right) + 10\left( {5\widehat I \times 2\widehat J} \right) + 3\widehat J \times \left( { - 4\widehat I} \right) + \left( {3\widehat J \times 2\widehat J} \right)\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two gyroscopes are made exactly alike except that the spinning disk in one is made of low-density aluminum, whereas the disk in the other is made of high-density lead. If they have the same spin angular speeds and the same torque is applied to both, which gyroscope processes faster?

The Bohr model currently predicts the main energy levels not only for atomic hydrogen but also for other “one-electron” atoms where all but one of the atomic electrons has been removed, such as in He+ (one electron removed) or (two electrons removed) Li++. (a) Predict the energy levels in for a system consisting of a nucleus containing protons and just one electron. You need no recapitulate the entire derivation for the Bohr model, but do explain the changes you have to make to take into account the factor . (b) The negative muon(μ-) behaves like a heavy electron, with the same charge as the electron but with a mass 207 times as large as the electron mass. As a moving μ- comes to rest in matter, it tends to knock electrons out of atoms and settle down onto a nucleus to form a “one-muon” atom. For a system consisting of a lead nucleus ( Pb208has 82 protons and 126 neutrons) and just one negative muon, predict the energy in of a photon emitted in a transition from the first excited state to the ground state. The high-energy photons emitted by transitions between energy levels in such “muonic atoms” are easily observed in experiments with muons. (c) Calculate the radius of the smallest Bohr orbit for a μ- bound to a lead nucleus ( Pb208has 82 protons and 126 neutrons). Compare with the approximate radius of the lead nucleus (remember that the radius of a proton or neutron is about 1×10-15m, and the nucleons are packed closely together in the nucleus).

Comments: This analysis in terms of the simple Bohr model hints at the results of a full quantum-mechanical analysis, which shows that in the ground state of the lead-muon system there is a rather high probability for finding the muon inside the lead nucleus. Nothing in quantum mechanics forbids this penetration, especially since the muon does not participate in the strong intersection. Electrons in an atom can also be found inside the nucleus, but the probability is very low, because on average the electrons are very far from the nucleus, unlike the muon.

The eventual fate of the μ- in a muonic atom is that it either decays into an electron, neutrino, and antineutrino, or it reacts through the weak interaction with a proton in the nucleus to produce a neutron and a neutrino. This “muon capture” reaction is more likely if the probability is high for the muon to be found inside the nucleus, as is the case with heavy nuclei such as lead.

A comet orbits the sun (figure). When it is at location 1 it is a distance d1from the sun. When the comet is at location 2, it is a distanced2from the sun, and has magnitude of momentump2. (a) when the comet is at location 1, what is the direction of LA? (b) when the comet is at location 1, what is magnitude of LA? (c) When the comet is at location 2, what is the direction of LA? (d) when the comet is at location 2, what is the magnitude ofLA ? Later we’ll see that the angular momentum principle tells us that the angular momentum at location 1 must be equal to the angular momentum at location 2.

A device consists of eight balls, each of massattached to the ends of low-mass spokes of length L so the radius of rotation of ball is L/2. The device is mounted in the vertical plane, as shown in Figure 11.73. The axle is help up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. A lump of clay with massm falls and sticks to one of the balls at the location shown, when the spoke attached to that ball is 45°to the horizontal. Just before the impact the clay has a speed v, and the wheel is rotating counter clock wise with angular speedω .

(a.) Which of the following statements are true about the device and the clay, for angular momentum relative to the axle of the device? (1) the angular momentum of the device + clay just after the collision is equal to the angular momentum of the device +clay just before the collision. (2) The angular momentum of the falling clay is zero because the clay is moving in a straight line. (3) Just before the collision, the angular momentum of the wheel is 0. (4) The angular momentum of the device is the sum of the angular momenta of all eight balls. (5) The angular momentum of the device is the same before and after the collision. (b) Just before the impact, what is the (vector) angular momentum of the combined system of device plus clay about the center C? (As usual, xis to the right, yis up, and zis out of the screen, toward you) (c) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center C? (d) Just after the impact, what is the (vector) angular velocity of the device? (e) Qualitatively. What happens to the total linear momentum is changed system? Why? (1) some of the linear momentum is changed into energy. (2) some of the linear momentum is changed into angular momentum. (3) There is no change because linear momentum is always conserved. (4) The downward linear momentum decreases because the axle exerts an upwards force. (f) qualitatively, what happens to the total kinetic energy of the combined system? Why? (1) some of the kinetic energy is changed into linear momentum. (2) some of the kinetic energy is changed into angular momentum. (3) The total kinetic energy decreases because there is an increase of internal energy in this inelastic collision. (4) There is no change because kinetic energy is always conserved.

The nucleus dysprosium-160 (containing 160 nucleons) acts like a spinning object with quantized Angular momentum. L2=lI+1h2, and for this nucleus it turns out thatmust be an even integer . When a Dy-160 nucleus drops from the l = 2 state to the l = 0 state, it emits an 87 KeV photon . (a) what is the moment of inertia of the Dy-160 nucleus? (b) Given your result from part (a), find the approximate radius of the Dy-160 nucleus, assuming it is spherical. (In fact, these and similar experimental observation have shown that some nuclei are not quite spherical.) (c) The radius of a (spherical) nucleus is given approximately by(1.3x10-15m)A13 , where A is the total number of protons and neutrons. Compare this prediction with your result in part (b).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free