The answer would be "the vector sum of forces acting on a particle or body."
Hope that helped ^^
The earth tilt about 23 degrees which leads to the formation of seasons on earth.
<h3>At what degree the earth tilt?</h3>
This is tilted at 23 degrees from the vertical. The northern hemisphere is in the winter season.
This is the path as the earth revolves around the sun. It takes 1 year to complete this path.
This is the path as the earth rotates around its.
So we can conclude that the earth tilt about 23 degrees which leads to the formation of seasons on earth.
Learn more about season here: brainly.com/question/25870256
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Answer:
0.339 kgm²
Explanation:
We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.
Since T = 2π√(I/mgh), making I subject of the formula, we have
I = mghT²/4π²
Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.
So, T = 241 s/113 = 2.133 s
So, Substituting the values of the variables into I, we have
I = mghT²/4π²
I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²
I = 15.63/4π² kgm²
I = 0.396 kgm²
Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis
I' = I - mh²
I' = 0.396 kgm² - 2.15 kg × (0.163 m)²
I' = 0.396 kgm² - 0.057 kgm²
I' = 0.339 kgm²
Answer:
x=0.46m, speed=7.9m/s
Explanation:
Using the concept of conservation of energy:
1. kinetic energy of mass m and velocity v: 
2. gravitational potential energy of mass m, grav. acc. g and height h: 
3. potential energy in a spring with spring constant k and displacement from equilibrium x: 
Calculating x:


Calculating the speed:



Solving for
:

Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm