No, this is false.
In principle, every well can run dry, and this includes a well that has reached the water table. If the water is taken from the well faster than it's replenishing itself, the well will run dry- and in this situation, typically people need to look for a new place to have a well.
<h2>
Answer:The angle between the refracted ray and the normal.</h2>
Explanation:
Refraction is defined as the phenomenon of bending or changing course of a light ray when it passes from one medium to another medium.
The new ray formed after the bending at the surface of new medium is called the refracted ray.
The angle between the refracted ray and the normal of the point on the surface of the new medium where the ray enters is called angle of refraction.
Refer the attachment for instance.
In short,it is the angle between the refracted ray and the normal.
Answer:
v(7) = 52.915 m/s
Explanation:
First, find the value for acceleration.
F = ma
100 = .5 * a
a = 200 m/s²
Next find the velocity at x = 7 using kinematic equations.
v² = v₀² + 2a(Δx)
v² = (0)² + 2(200)(7)
v = 
v = 52.915 m/s
D=-5m
a(gravity)=-9.8m/s^2
vi= 0m/s
t=?
use equation d=vi*t+0.5a*t^2
because vi=0, you can cross out vi*t because anything multiplied by 0= 0
rearrange the equation to say t^2=d/0.5a
t^2= -5/-4.9
t^2=1.02
find the square root...
final answer: t=1s
Answer:
dg= 942m
Explanation:
given the depth of the granite Us dg = 500m
time between the explosion t = 0.99s
the speed of sound in granite is Vg = 6000m/s
First of all calculate the time it takes the sound waves to travel down through the lake
Vw = dw/t1
t1 = dw/Vw
t1 = 500/1480
t1 = 0.338s.
Let dg be the depth of the granite basin, so the time it takes for the sound to travel down through the granite is t2 = dg/6000m/s......equation(1)
So the total time it takes to travel down to the oil surface will be
t1/2 = t1 + t2
t1/2= 0.338 + dg/6000.
since the reflection on the oil does not change the speed of sound, the sound will take travelling upto the surface the same time it takes to reach the oil
so; t = 2 t1/2
t1/2 = t/2 = 0.99s/2 = 0.495
Now insert into the values of t1/2 into the equation (1) and solve for dg;
we get 0.495 = 0.338 + dg/6000
dg = (0.495 - 0.338) x 6000
dg = 942m.
