I need help with this assignment? - key running toy car
1. You accidentally throw your car keys horizontally at 6.0 m / s from a cliff 66 meters high. How far from the base of the cliff is the key to get?
2.A toy car along the edge of a table that is 1,150 m high. The car landed 0385 ft from the base of the table.
(a) How long will it take the car to fall?
(b) How fast was the car rolls off the table?
3.You start-up platform diving. Which was 2.4 m / s, 2.9 s and then water. What was the platform and at what distance from the edge of the platform will hit the water? Neglect air resistance.
4. A ball is hit by a degree of dough into a 50 ° angle and allows the closure of the external field, 97 meters away. Let us assume that the fence the same height as the height and find the speedING ball when it left the bat. Neglect air resistance.
Key Running Toy Car I Need Help With This Assignment?
5 comments:
1. Dy = 1 / 2 A * t ^ 2 + Vy * t, V = 0 (velocity is horizontal)
t = sqrt (66m / (1 / 2 * 9.8 m / s ^ 2)) = 3.67 s.
Dx = Vx * t = 6 m / s * 3.67 s = 22 m.
2. (A) dy = 1 / 2 A * t ^ 2 + Vy * t, V = 0
t = sqrt (1.150m / (1 / 2 * g)) = 0,484 s
(b) dx = Vx * t, so that Vx = dx / t = 0.385m/0.484s = 0.795 m / s
3. Dy = 1 / 2 A * t ^ 2 + Vy * t, V = 0
and D = 1 / 2 * 9.8 m / s * (2.9s up are) ^ 2 = 41.2 m
Dx = Vx * t = 2.4 m / s * 2.9 s = 6,96 m
4. Dy = 0, Dx = 97 m, a = 50 degrees
Dx = Vx * t = V * cos (a) * t
t = dx / (V * cos (a))
-1 / 2 * g * t ^ 2 + Dy Vyt = = 0 (V = V * sin (a))
Division by t, -1 / 2 * g * t + V * sin (a) = 0, or
t = V * sin (a) / (1 / 2 * g)). Equalizing the two expressions t:
V * sin (a) / (1 / 2 * g) = dx / (V * cos (a)), we multiply both sides by V
V ^ 2 * sin (a) / (1 / 2 * g) = dx /cos (a)
V = sqrt (dx / cos (a) / sin (a) * (1 / 2 * g))
V = 31.1 m / s
Did you ever decide class since the beginning of the semester, or just beginning, before one visit to this job?
u have to do yourself to mate
If you are not really a problem of his task, I recommend you leave the course.
Problem number 1:
First, how long it would take to reach the ground 66. Changes in the initial position by time equals speed) more than 1 / 2 times accelartion (time squared. Is Note that) vertically for a falling object, first (zero speed, so that in the event of a change in the position equal to 1 / 2 times the acceleration of time is squared. Clearing of the time. Well, how long should the keys in the air, and you know how fast the keys to move horizontally to be able to decrypt the key action on the basis of time to go speed.
Problem 2:
You should be able to see part A in the same way that he found the time to drop the key to the problemone. For Part B, which usually comes the second part of Task 1 in the opposite direction. Use the same equation, but now you know the time and distance to solve speed.
Problem 3:
To begin with, now is the time to fall and wants to know the extent to which the position is. Use the same equation to solve since the beginning of Part A, and only in part-time position at INSEAD. The portion of the horizontal distance is the same as the end of task 1
I do not remember exactly how to solve the problem 4, but should be able to do the same equations as the last three.
Although it could take place, and said that the answers should be based on all of these things if you want to go a chance to the class, try everything possible throughProblems. Even if someone answers, please try here to solve problems themselves!
Almond9090 - You know, you really do any favors for this guy his homework for him / her.
They can help in this task to know the constant acceleration formulas. These are basic questions, if you know these formulas, as well as knowledge in the resolution of vectors into components.
If you must use to do your homework, then they should be doing physics.
Post a Comment