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".
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Answer:
Explanation:
The formula for the potential energy of a dipole placed in an electric field is given by
U = - pE Cos θ
where, θ is the angle between dipole moment and the electric field vector.
For θ = 0°,
initial potential energy, Ui = - pE
For θ = 180°,
final potential energy, Uf = - pE Cos 180 = pE
Change in potential energy
ΔU = Uf - Ui
ΔU = pE - (-pE)
ΔU = 2pE
Answer:
s = 1.7 m
Explanation:
from the question we are given the following:
Mass of package (m) = 5 kg
mass of the asteriod (M) = 7.6 x 10^{20} kg
radius = 8 x 10^5 m
velocity of package (v) = 170 m/s
spring constant (k) = 2.8 N/m
compression (s) = ?
Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore
• Ei = Ef
• Ei = energy in the spring + gravitational potential energy of the system
• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}
• Ef = kinetic energy of the object
• Ef = \frac{1}{2}mv^{2}
• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}
• s = 
s = 
s = 1.7 m
