Answer: 4575N
Explanation:
For y component, W = mgcosø
W = 500×9.8cos21
W = 4574.54N
Find the diagram in the attached file
<span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400
This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.
Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.
Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600
This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.
Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)
The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.
x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>
Answer:
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Answer:
Explanation:
the light ray leaving a medium in contrast to the entering or incident ray.
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
