A neon sign _________ light. A. emits B. filters C. reflects D. transmits
1 answer:
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Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:
B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º
C) The maximum force is: F = 24 N
D) The time to stop the block is: t = 25 s
Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.
B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.
x-axis
-10 + 12 sin 60 + Fₓ = 0
Fₓ = 10- 12 sin 60 = -0.39 N
y-axis
12 cos 60 - 6 + F_y = 0
F_y = 6 - 12 cos 60 = 0 N
We can give the result of the force in two ways:
- Form of coordinates F = -0.39 i ^ N
- Form of module and angle.
Let's use Pythagoras' theorem to find the modulus.
F = 0.39N
We use trigonometry for the angle.
tan θ= 0º
The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.
θ = 180 + θ'
θ = 180 + 0
θ = 180º
C) if the three forces can be moved and the maximum force occurs when they are all linear.
10+ 6 + 6 + F = 0
F = -24 N
D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.
a =
a = 4 m / s²
The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s
If we use kinematic relations.
v = v₀ - a t
Final velocity when stopped is zero
t =
t = 100/4
t = 25 s
In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:
B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º
C) The maximum force is: F = 24 N
D) The time to stop the block is: t = 25 s
Learn more about Newton's second law here: brainly.com/question/25545050
