Hi there!
We can use the following (derived) equation to solve for the final velocity given height:
vf = √2gh
We can rearrange to solve for height:
vf² = 2gh
vf²/2g = h
Plug in the given values (g = 9.81 m/s²)
(13)²/2(9.81) = 8.614 m
We can calculate time using the equation:
vf = vi + at, where:
vi = initial velocity (since dropped from rest, = 0 m/s)
a = acceleration (in this instance, due to gravity)
Plug in values:
13 = at
13/a = t
13/9.81 = 1.325 sec
Answer:
<u>We are given:</u>
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
<u />
<u>Solving for Displacement:</u>
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
Assuming that the angle is the same for both ropes, then D. is the answer. You have to consider also if the ropes are close together or far apart and if the force to move the object is in line with the ropes or perpendicular to them.
<span />
All of the following are non-renewable resources except
O natural gas
O oil
O minerals
O <em>water ✓ </em>
- <em>Water </em><em>is </em><em>a </em><em>renewable </em><em>source </em><em>because </em><em>evaporation </em><em>and </em><em>condensation </em><em>takes </em><em>place </em><em>everytime </em><em>on </em><em>our </em><em>planet</em>
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h =
h =
h = 1.55 m