This is the period in a simple harmonic motion which is 2 seconds in this question.
<h3>
What is Period ?</h3>
The period of an oscillatory object can be defined as the total time taken by a vibrating body to make one complete revolution about a reference point.
We are given the below question
2×3.14√(1.0m/(9.8〖ms〗^(2) )= T
This question can as well be expressed as
2π√(L/g) which is equal to period T.
In a nut shell, Period T = 2×3.14√(1.0m/9.8)
T = 6.28√0.102
T = 6.28 × 0.32
T = 2.006 s
Therefore, the period T of the oscillation is 2 seconds approximately.
Learn more about Period here: brainly.com/question/12588483
#SPJ1
Answer:
The car C has KE = 100, PE = 0
Explanation:
The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.
The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.
Kinetic energy is the energy that a body possesses due to its motion.
Potential energy is the energy a body possesses due to its position.
From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.
In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.
Therefore, the car C has KE = 100, PE = 0
As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second
44.0 m/s - 2.0 m/s = 42.0 m/s
42.0 m/s / 12 s = 3.5 m/s2
the acceleration of the rock would be 3.5 m/s2
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
