Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s
Answer:
D. mass to see how it affected stretch length of a rubber band
Ranboo oobnar have a good day
Answer: 1.88 N
Explanation:
Data:
Force = 4.00N
angle = 62°
horizontal force = ?
Solution:
The trigonometric ratio that relates horizontal - leg to hypotenuse is the cosine.
That ratio is:
horizontal - leg
cos(angle) = -------------------------
hypotenuse
So, applied to the force, that is:
horizontal force
cos (angle) = -----------------------------------
total force
So, clearing the horizontal component you get:
horizontal force = force * cos (angle)
Substitute the data given:
horizontal force = 4.00N * cos(62°) = 4.00N * 0.4695 = 1.88 N
Answer: 1.88N
