The period of the oscillations.T = 1.2042s
Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.
Typically, the mathematical formula for the moment of inertia is
T = 2 π √(I / mgd)
Therefore, a moment of inertia
I = 9.00×10-3 + md^2 ;
I=9.00*10^{-3}+ 0.5 * 0.3^2
I=0.054
T=2
T=1.2042s
The period of the oscillations.T = 1.2042s
Read more about the period of the oscillations. brainly.com/question/14394641
#SPJ1
Answer:
a) 141.6m
b) 8.4m/s
Explanation:
a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:
hence, the total distance is 141.6m
b) the mean velocity of the total trajectory is given by:
hence, the mean velocity is 8.4 m/s
Length of the pipe = 0.39 m
Number of harmonics = 3
Now there are 3 loops so here we can say
now here at the center of the pipe it will form Node
we need to find the distance of nearest antinode
So distance between node and its nearest antinode will be
So the distance will be 6.5 cm
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
