A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release
1 answer:
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The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.
The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.
The change in momentum of the system of the carts is 0.
<h3>Initial momentum of the carts before collision</h3>The momentum of the carts before the collision is calculated as follows;
P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s
P(blue) = 1.5 x 0 = 0
<h3>Momentum of the carts after collision</h3>The momentum of the carts after the collision is calculated as follows;
P(red) = 0.5 x 0.1 = 0.05 kgm/s
P(blue) = 1.5 0.1 = 0.15 kgm/s
<h3>Change in momentum of the carts</h3>ΔP = (0.05 + 0.15) - (0.2)
ΔP = 0
Learn more about momentum here: brainly.com/question/7538238
Answer:
The magnitude of the resultant displacement is 21 mi and its direction is 16.7° north of west
Explanation:
Hi there!
Please see the figure for a better understanding of the problem. The total displacement vector will be the sum of both displacements:
The vector for the first displacement is:
First displacement = (20 mi, 0)
The second displacement:
Second displacement = (0, 6.0 mi)
The resultant displacement will be:
R = (20 mi, 0) + (0, 6.0 mi) = (20 mi + 0, 0 + 6.0 mi) = (20 mi, 6.0 mi)
The magnitude of this vector will be:
The magnitude of the vector displacement is 21 mi.
To find the direction of the vector R, we have to apply trigonometry:
In a right triangle the following trigonometric rule applies:
cos θ = adjacent side to the angle/ hypotenuse
In this case:
cos θ = 20 mi / magnitude of R
θ = 16.7°
The direction of the vector is 16.7° north of west.
