Answer:
a) I = (
+ 
) L² , b) w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
Explanation:
a) The moment of inertia is a scalar that represents the inertia in circular motion, therefore it is an additive quantity.
The moment of inertia of a rod held at one end is
I₁ = 1/3 M L²
The moment of inertia of the mass at y = L
I₂ = m y²
The total inertia method
I = I₁ + I₂
I = \frac{1}{3} M L² + m (\frac{2}{3} L)²
I = ( + ) L²
b) The conservation of angular momentum, where the system is formed by the masses and the bar, in such a way that all the forces during the collision are internal.
Initial instant. Before the crash
L₀ = I₂ w₀
angular and linear velocity are related
w₀ = y v₀
w₀ = 
L v₀
L₀ = I₂ y v₀
Final moment. After the crash
= I w
how angular momentum is conserved
L₀ = L_{f}
I₂ y v₀ = I w
substitute
m (
)² (\frac{2L}{3} v₀ = ( + ) L² w
m L³ v₀ = ( + ) L² w
m L v₀ = ( + ) w
L v₀ = 
w
w = (\frac{27 M}{18 m} + 2)⁻¹ Lv₀
Answer: 5.075Ns
Explanation:
Given the following :
Mass of ball = 145g
Initial Speed of ball = 15m/s
Final speed of ball when hit by the batter = - 20m/s ( Opposite direction)
The impulse of a body is represented using the relation:
Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))
Therefore, using:
m(v - u) = impulse
Mass of ball = 145 / 1000 = 0.145kg
Impulse = 0.145(- 20 - 15)
Impulse = 0.145(-35)
Impulse = 5.075Ns
should be lifespan dovelopment
Answer:
Explanation:
For lens A
object distance u = - 13.1 cm , focal length f = 6.19 cm
From lens formula
1/v - 1/u = 1/f
1 / v + 1/13.1 = 1/6.19
1/v = 1/6.19 - 1/13.1
= .16155 - .07633
= .08522
v = 11.7 3 cm
For lens B
object distance u = - ( 55.7 - 11.73) = - 43.97 cm , focal length f = 27.9 cm
From lens formula
1/v - 1/u = 1/f
1 / v + 1/43.97 = 1/27.9
1/v = 1/27.9 - 1/43.97
= .03584 - .022742
= .013098
v = 76.35 cm
Image will be formed 76.35 cm behind lens B .
magnification of lens system
= m₁ x m₂ , m₁ is magnification by lens A and m₂ is magnification by lens B
= (11.73 / 13.1) x (76.35 / 43.97)
= .8954 x 1.73
= 1.5547
size of image = total magnification x size of object
= 1.5547 x 6.47
= 10 cm approx. The first image will be real and inverted and second image will be erect with respect to object.
Answer:
a)6.7m/S
b)6.8m/s
Explanation:
Hello ! To solve the point b you must follow the steps below
1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)
2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)
3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)
4)
the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation
where
Vfy= final verticaly speed
Vy=initial verticaly speed=0.59m/S
g=gravity=9.81m/S^2
y=height of slide=2.31m
solving
The last step is to add the velocity components vectorally at the end of the route with the following equation
point A
taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.
vy=0
Vfy=
=6.7m/s
