1. Your new designer chair has an S-shaped tubular metal frame that behaves just like a spring. When your friend, who weighs 600
1 answer:
Explanation:
Given that,
Weight of the friend, W = 600 N
When the friend sits on the metal frame it bends downward 4 cm, we can say that the compression in the it is 4 cm or 0.04 m
To find,
Spring constant for this chair or k
Solve :
The weight of an object is equal to the force exerted by the gravitational force, F = 600 N
According to Hooke's law, the force exerted by the spring is given by :
F = kx
k is the spring constant
k = 15000 N/m
Therefore, the spring constant of the spring is 15000 N/m.
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Inverse square law: 
where
is the intensity at distance 1
is the intensity at distance 2
is distance 1
is distance 2
The inverse squared law state that intensity decreases in inverse proportion to the distance squared. So if light obeyed that rule, it will decreases its intensity as the square of the distance increases.
We can conclude that the correct answer is: true.
where
The inverse squared law state that intensity decreases in inverse proportion to the distance squared. So if light obeyed that rule, it will decreases its intensity as the square of the distance increases.
We can conclude that the correct answer is: true.
Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis
KE = 0.23 J
given parameters
- Disk radius R = 15 cm = 0.15 m
- Cylinder radius r = 1.5 cm = 0.0015 m
- Disk mass M = 20 kg
- Time t = 1.2 s
- Force F = 12 N
to find
- Kinetic energy (KE)
This exercise must be solved in parts:
1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation
KE = ½ I w²
Where KE is the kinetic energy, I the moment of inertia and w the angular velocity
The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is
I_{ total} = I_{ disk} + I_{ cylinder}
the moments of inertia with respect to an axis passing through the center of mass are tabulated
disk I_{disk} = ½ M R²
cylinder I_{cylinder} = ½ m r²
where M and m are the masses of the disk and cylinder respectively, R and r their radii
I_{total} = ½ (M R² + m r²) = ½ M R² ( )
I_{total} = ½ M R² ( ) =
M R² (1 + 0.005 m)
As the shaft mass is much lighter than the disk mass , the last term is very small, which is why we despise it.
I_{total} = ½ M R²
2nd part. Let's use Newton's second law for endowment motion
τ = I α
α = l
τ = F R
α =
With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)
w = w₀ + α t
where w is the angular velocity, alpha is the angular acceleration and t is the time
w = 0 +
we substitute in the kinetic energy equation
KE = ½ I_{total} ( )²
KE = ½
let's substitute
KE =
KE = F² R² t² / M
let's calculate
KE = 12² 0.15² 1.2² / 20
KE = 0.23 J
With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is
KE = 0.23 j
learn more about rotational kinetic energy here: