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
Answer:
330.5 m
Explanation:
In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .
The maximum height will be calculated as;
where ∝ is the angle of launch = 30°
vi= initial launch velocity = 40 m/s
g= 10 m/s²
h= 40²*sin²40° / 2*10
h={1600*0.4132 }/ 20
h= 661.1/2 = 330.5 m
Answer:a. 24 kg m/s
b. 3/5s
Explanation:
a.impulse is the change in momentum so at first the momentum is zero because the ball was at rest and the final momentum is 1.2kg*20m/s=24 kg m/s
so the impulse would be (24-0) kg m/s=24 kg m/s
b. so the impulse equation is impulse is force *delts time
so 24 kg m/s=40N*t
t=24 kg m/s /40N=3/5 s
Answer:
109.32 N/m
Explanation:
Given that
Mass of the hung object, m = 8 kg
Period of oscillation of object, T = 1.7 s
Force constant, k = ?
Recall that the period of oscillation of a Simple Harmonic Motion is given as
T = 2π √(m/k), where
T = period of oscillation
m = mass of object and
k = force constant if the spring
Since we are looking for the force constant, if we make "k" the subject of the formula, we have
k = 4π²m / T², now we go ahead to substitute our given values from the question
k = (4 * π² * 8) / 1.7²
k = 315.91 / 2.89
k = 109.32 N/m
Therefore, the force constant of the spring is 109.32 N/m
