Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
As we know that speed of package is same as that of helicopter in horizontal direction
So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall
So we have



Part b)
Distance from helicopter is same as the distance of free fall
so we will have

Part c)
If helicopter is rising upwards with uniform speed
then final speed of the package after time t is given as


Part d)
distance from helicopter

Answer:
(i) 208 cm from the pivot
(ii) Move further from the pivot
Explanation:
(i) Sum of the moments about the pivot of the seesaw is zero.
∑τ = Iα
(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0
1250 Nm + 600 N x = 0
x = -2.08 m
Kenny should sit 208 cm on the other side of the pivot.
(ii) To increase the torque, Kenny should move away from the pivot.
Answer:
a)106.48 x 10⁵ kg.m²
b)144.97 x 10⁵ kgm² s⁻¹
Explanation:
a)Given
m = 5500 kg
l = 44 m
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Putting all the required values, moment of inertia of one blade will be
= 1/3 x 5500 x 44²
= 35.49 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 35.49 x 10⁵ kg.m²
= 106.48 x 10⁵ kg.m²
b) Angular momentum 'L' is given by
L =
x ω
where,
= moment of inertia of turbine i.e 106.48 x 10⁵ kg.m²
ω=angular velocity =2π f
f is frequency of rotation of blade i.e 13 rpm
f = 13 rpm=>= 13 / 60 revolution per second
ω = 2π f => 2π x 13 / 60 rad / s
L=
x ω =>106.48 x 10⁵ x 2π x 13 / 60
= 144.97 x 10⁵ kgm² s⁻¹
Answer:
v₀ₓ = 15 m / s,
= 5.2 m / s
v = 15.87 m / s
, θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =
² - 2 g y
² =
² + 2 g y
= √ (
² + 2 gy
Let's calculate
= √ (1.25² + 2 9.8 1.3)
= √ (27.04)
= 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² +
²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
= 5.2 m / s
Explanation:
3
i believe that they are all going at 3.2 meters each, I did 4 times 0.8