Answered: What are the magnitude and direction of… | bartleby

Related questions

Question

What are the magnitude and direction of the cross product rAB×F, where rAB is the position vector and F is the force vector, respectively?

В
20°f
YAB d
60°
F
A

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

Scalars and vectors: Vector A has a magnitude of 9.0 and Vector B has a magnitude of 3.0. If the vectors are at an angle of 30.0º, what is the magnitude of the cross product A x B? Here are the choices: 13.5 16.2 23.4 27.0
Using the definition of dot product: A B = ABCOS (0AB) = AxBx + Ay By + A₂B₂ Find the angle between the following vectors: (c) A = 1î + 3j+0k B = 3î + 1) + Ok A = 1î - 3j + 2k B = -31 + 1) + 0k (b) (d) A = 1î + 1ĵ + Ok B = 21-3j+0k A = 21-5j-1k B = 3î + 11 + 3k
Two vectors are given by a = 1.5î−4.0ĵ and b = −11.9î + 8.1ĵ. What is the magnitude of b? i already found out magnitude of b is 1.44 but how to calculate the angle between vector b and the positive x-axis?
6. Vector A has a magnitude of 5.00 units, and vector B has a magnitude of 9.00 units. The two vectors make an angle of 50.0° with cach other. Find A B. Note: In Problems 7 and 8, calculate numerical answers to three significant figures as usual. 7. Find the scalar product of the vectors in Figure P7.7. 118 32.8 N 132° 17.3 cm Figure P7.7 8. Using the definition of the scalar product, find the angles between (a) A = 3i-2j and B = 4i-4j. (b) A= -2i + 4j and B= 3i- 4j + 2k, and (c) A=i-2j + 2k and B = 3j + 4k. SECTION 7.4 Work Done by a Varying Force 9. A particle is subject to a force F that varies with position as shown in Figure P7.9. Find the work done by the force on the particle as it moves (a) from x = 0 to x = 5.00 m, (b) from x = 5.00 m to x = 10.0 m, and (c) from x= 10.0 m to x = 15.0 m. (d) What is the total work done by the force over the distance x= 0 tox= 15.0 m: 7, (N) X (m) 121 10 12 14 16 Figure P7.9 Problems 9 andl 22.
Two vectors have a dot product of 7 m² and a cross product of magnitude 4.95 m². What is the angle between the two vectors?
1. The vector a has a magnitude of 5.00 units and the vector ba magnitude of 7.00 units. If the angle between the vectors is 53.00, find their scalar product or dot product.
Two vectors have the following magnitude, A = 8.7 m and B = 10.6 m. Their vector product is: A:B = -2.7 m i+ 8.6 m k. What is the angle (in degrees) between the vectors A and B? Hint: Use |A:B| = AB sin(0) Note: Bold face letters represents a vector.
Use the definition of scalar product, a = ab cos 0, and the fact that a . the two vectors given by a = 3.01 +3.0 + 3.0k and b Number i Units = axbx + ab + a₂b₂ to calculate the angle between 4.0î + 9.0ĵ + 7.0k. =
Find a unit vector perpendicular to A = (î+ ĵ – Îk) and B = ( 2i + j- 3k). (Hints. One method to find the unit vector C perpendicular to A and B will be to use the fact that the cross product is perpendicular to both vectors A and B, but keep in mind that the magnitude of the unit vector is one. There is another method which is a little bit longer by using the fact that the dot product AC=0 and BC=0 where C is the unit vector )
For the vectors shown in the figure, find the magnitude and direction of the vector product A C. B' 6. 70° 40° 8. 48, directed out of the page 45. directed out of the page O 16. directed out of the page O 48, directed into the page O 45, directed into the page ) 16, directed into the page
For the vectors in the figure, with a = 8.0, b = 6.0, and c = 10 what are the magnitude and the direction b×c?
What is the dot product of two vectors A = 2x + 4 y + 7z and B = -3x + 8y - 2 z where it is understood that A and B are 3 dimensional vectors, and x, y, and z are vectors of length 1 along the x, y, and z axes. A scalar +12 A scalar +16 A vector perpendicular to both A and B A vector 6x +32y -14z