Jeremy is walking through the woods, marking diseased trees for removal. He walks for 2.44 km, at a heading of 13.9° west of nor
1 answer:
We conclude that Jeremy walked 2.37km North and 0.59 km West.
<h3>How to get the North and West components?</h3>
We can assume that the distance that he walked is the hypotenuse of a right triangle, like the one you can see in the image below.
There, you can see that the adjacent cathetus would be the displacement in the North direction, while the opposite cathetus would be the displacement in the West direction.
Then we can use the relations.
- Sin(θ) = (opposite cathetus)/(hypotenuse)
- Cos(θ) = (adacent cathetus)/(hypotenuse).
With θ = 13.9° and hypotenuse = 2.44km
Solving these two we will get:
sin(13.9°) = Y/2.44km
sin(13.9°)*2.44km = Y = 0.59 km
cos(13.9°) = X/2.44km
cos(13.9°)*2.44km = X = 2.37km
So we can conclude that he walked 2.37km North and 0.59 km West.
If you want to learn more about right triangles, you can read:
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Answer:
100N
Explanation:
Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.
The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.
Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.
That is the force the boy exert on the man during the shove.
The tension in the string B) It quadruples.
Explanation:
The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:
where:
T is the tension in the string
m is the mass of the ball
v is the speed of the ball
r is the radius of the circle (the lenght of the string)
In this problem, we are told that the speed of the ball is doubled, so
v' = 2v
Substituting into the previous equation, we find the new tension in the string:
Therefore, the tension in the string will quadruple.
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