Describe how switching the desk lamp on and off shows that light waves transfer energy
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Answer:
F_ab = 260.17 N
Explanation:
Given:
- Moment of force F about CD, (M)_cd = 57 Nm
Find:
- First we will write down the position vectors of points A, B , C , D:
- We will take left and bottom most corner of cube to be the origin.
- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.
- The position vectors wrt to the origin are:
Point A = 0.2 k
Point B = 0.4 i + 0.2 j
Point C = 0.2 j + 0.4 k
Point D = 0.4 i + 0.4 k
- Now we will determine the Force vector F_ab along vector AB.
vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k
unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)
= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
Hence,
vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
- Now, form a unit vector along the line CD:
vec(CD) = D - C = 0.4 i - 0.2 j
unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)
= [sqrt(5)]*(0.4 i - 0.2 j)
- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:
vec(AC) =r_ac = C - A = 0.2 j + 0.2 k
- Now the moment about a line CD due to force is:
(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )
The cross product of r_ac and vec(F_ab) is as follows:
(M)_c = ( r_ac x vec(F_ab) ) :
(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]
The dot product of (M)_c and unit (CD) is as follows:
(M)_cd = unit(CD) . (M)_c :
(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)
(M)_cd = F_ab*(sqrt(30) / 25)
- The given magnitude of the moment is (M)_cd. Calculate F_ab:
57 = F_ab*(sqrt(30) / 25)
F_ab = 260.17 N
The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:
x = 0.135 cm
Given parameters
- The mass m = 220 g = 0.220 kg
- The spring cosntnate3 k = 7.0 N / m
- Initial displacement A = 5.2 cm = 5.2 10-2 m
To find
- The position where the kinetic and potential energy are equal
A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.
x = A cos wt + fi
w² =
Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.
The speed is defined by the variation of the position with respect to time.
v =
let's evaluate
v = - A w sin (wt + Ф)
Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.
0 = - A w sin Ф
For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero
x = A cos wt
Let's find the point where the kinetic and potential energy are equal.
K = U
½ m v² = m g x
we substitute
½ A² w² sin² wt = g A cos wt
sin² wt = cos wt
let's calculate
w =
w = 5.64 rad / s
sin² 5.64t = 2 9.8 / 0.052 cos 5.64t
sin² 5.64t = 376.92 cos 5.64 t
1 - cos² 5.64t = 376.92 cos 5.64t
cos² 5.64t -376.92 cos564t -1 = 0
we make the change of variable
x = cos 5.64t
x²- 376.92 x - 1 = 0
x = 0.026
cos 5.64t = 0.026
Let's find the displacement for this time
x = 5.2 10-2 0.026
x = 1.35 10-3 m
In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:
x = 0.135 cm
Learn more here: brainly.com/question/15707891