Groundwater is the water found underground in the cracks and spaces in soil, sand and rock. It is stored in and moves slowly through geologic formations of soil, sand and rocks called aquifers.
Hope this helps.
Please mark as brainliest..........
As stated in the statement, we will apply energy conservation to solve this problem.
From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as
Where,
m = mass
= initial and final velocity
g = Gravity
h = height
As the mass is tHe same and the final height is zero we have that the expression is now:
Here is the rule for see-saws here on Earth, and there is no reason
to expect that it doesn't work exactly the same anywhere else:
(weight) x (distance from the pivot) <u>on one side</u>
is equal to
(weight) x (distance from the pivot) <u>on the other side</u>.
That's why, when Dad and Tiny Tommy get on the see-saw, Dad sits
closer to the pivot and Tiny Tommy sits farther away from it.
(Dad's weight) x (short length) = (Tiny Tommy's weight) x (longer length).
So now we come to the strange beings on the alien planet.
There are three choices right away that both work:
<u>#1).</u>
(400 N) in the middle-seat, facing (200 N) in the end-seat.
(400) x (1) = (200) x (2)
<u>#2).</u>
(200 N) in the middle-seat, facing (100 N) in the end-seat.
(200) x (1) = (100) x (2)
<u>#3).</u>
On one side: (300 N) in the end-seat (300) x (2) = <u>600</u>
On the other side:
(400 N) in the middle-seat (400) x (1) = 400
and (100 N) in the end-seat (100) x (2) = 200
Total . . . . . . . . . . . . <u>600</u>
These are the only ones to be identified at Harvard . . . . . . .
There may be many others but they haven't been discarvard.
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m
M= gpe / gh
G= gpe / mh
H=gpe / mg
