The angle between the two vectors is 126° 52' 11".
The given parameters;
vector A = 3.00i + 1.00j
vector B = 1.00i + 3.00j
The angle between the two vectors is calculated as follows;
The dot product of vectors A and B is calculated as;
A.B = ( 3i + 1j ) . ( 1i +3j )
= ( 3 × 1 ) + ( 1 × 3 )
= 3 + 3
= 6
The magnitude of vectors A and B is calculated as;
|A| = ![\sqrt[]{3^2 + 1^2} = \sqrt[]{10} \\](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B3%5E2%20%2B%201%5E2%7D%20%3D%20%5Csqrt%5B%5D%7B10%7D%20%5C%5C)
|B| = ![\sqrt[]{1^2 + 3^2} = \sqrt[]{10} \\](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B1%5E2%20%2B%203%5E2%7D%20%3D%20%5Csqrt%5B%5D%7B10%7D%20%5C%5C)
The angle between in two vectors is calculated as;
![Cos \theta = \frac{6}{\sqrt[]{10}\sqrt[]{10} } \\\\Cos \theta = \frac{6}{\ 10 }\\\\Cos \theta = 0.6\\\\\theta = cos^-^1 (0.6)\\\\\theta = 126\\](https://tex.z-dn.net/?f=Cos%20%5Ctheta%20%3D%20%5Cfrac%7B6%7D%7B%5Csqrt%5B%5D%7B10%7D%5Csqrt%5B%5D%7B10%7D%20%20%7D%20%5C%5C%5C%5CCos%20%5Ctheta%20%3D%20%5Cfrac%7B6%7D%7B%5C%2010%20%7D%5C%5C%5C%5CCos%20%5Ctheta%20%3D%200.6%5C%5C%5C%5C%5Ctheta%20%3D%20cos%5E-%5E1%20%280.6%29%5C%5C%5C%5C%5Ctheta%20%3D%20126%5C%5C)
Therefore, the angle between the two vectors is 126° 52' 11".
Learn more about vectors here:
brainly.com/question/25705666
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The answer would be a positive charge
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
Answer:
12.6 cm
Explanation:
We can use the mirror equation to find the distance of the image from the mirror:
where here we have
f = 9.50 cm is the focal length
p = 39 cm is the distance of the object from the mirror
Solving the equation for q, we find:
