Answer:
a. The laser tracking mechanism experiences a changing tangential velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration
Explanation:
a. The laser tracking mechanism experiences a changing tangential velocity
This is because the tangential velocity v = rω where r = radius of disc and ω = angular speed of discs. Since r is constant, v ∝ ω.
Since the angular speed changes from 200 rpm to 500 rpm, thus, the tangential velocity would also change.
So, the laser tracking mechanism experiences a changing tangential velocity
c. The laser tracking mechanism experiences a non-zero angular acceleration
Since angular acceleration, α = Δω/Δt where Δω = change in angular speed and Δt = change in time.
Since there is a change in angular speed from 200 rpm to 500 rpm in time Δt, there is thus a non-zero angular acceleration.
So, The laser tracking mechanism experiences a non-zero angular acceleration
d. The laser tracking mechanism experiences a non-zero tangential acceleration
Since tangential acceleration, a = rα where r = radius of disc and α = angular acceleration.
Since there is an angular acceleration of the disc, there is thus going to be a tangential acceleration given by a = rα.
So, the laser tracking mechanism experiences a non-zero tangential acceleration
Statement b is false because, the disc experiences a changing angular speed from 200 rpm to 500 rpm.
A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.
Answer:
a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N
Explanation:
For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.
Let's use trigonometry to break down the weight
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Wₓ = 1200 sin 30 = 600 N
W_y = 1200 cos 30 = 1039.23 N
Y axis
N- W_y = 0
N = W_y = 1039.23 N
Remember that the friction force always opposes the movement
a) in this case, the system will begin to move upwards, which is why friction is static
P -Wₓ -fr = 0
P = Wₓ + fr
as the system is moving the friction coefficient is dynamic
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = 600+ 207.85
P = 807.85 N
b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static
P + fr -Wx = 0
fr = μ N
fr = 0.20 1039.23
fr = 207.85 N
we substitute
P = Wₓ -fr
P = 600 - 207,846
P = 392.15 N
c) as the movement is continuous, the friction coefficient is dynamic
P - Wₓ + fr = 0
P = Wₓ - fr
fr = 0.15 1039.23
fr = 155.88 N
P = 600 - 155.88
P = 444.12 N
Which of the following best describes what occurs in a fission reaction?
A.
Two low mass nuclei are joined to form one nucleus.
B.
Electrons are shared between the nuclei.
C.
A single nucleus divides into two or more nuclei and gives off energy.
D.
A chemical reaction occurs between the nuclei.
Answer:
C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.
Answer:
C.
A single nucleus divides into two or more nuclei and gives off energy.
hope it is helpful to you
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
2(A + B)
15. The resultant of A and B is perpendicular to A
What is the angle between A and B?
(a) cos
(b) cos
La
(c) sin
(d) sin
Answer:
θ = cos^(-1) (-A/B)
Explanation:
The image of the reauktant forces A & B are missing, so i have attached it.
Now, from the attached image, we will see that;
Angle between A and B is θ
Also;
A = Bcos(180° − θ)
Now, in trigonometry, we know that;
cos(180° − θ) = -cosθ
Thus;
A = -Bcosθ
cosθ = -A/B
Thus;
θ = cos^(-1) (-A/B)
When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
Explanation:
energy of photon having wavelength of 400 nm = 1237.5/400 eV
= 3.1 eV.
Maximum kinetic energy of photoelectrons = 1.1 eV .
Threshold energy Ф = 3.1 - 1.1 = 2 eV .
energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .
Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy
= 4 - 2 = 2 eV .
Required kinetic energy K₀= 2 eV.
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
A radioactive material produces 1160 decays per minute at one time, and 4.0 h later produces 170 decays per minute. whats the half life
Answer:
Half life is 3.23 hours
Explanation:
Given
Decay rate at starting = 1160 decays per minute
Decay rate after 4 hours = 170 decays per minute
As we know know
[tex]N = N_0 *e ^{\Lambda *T}[/tex]
Substituting the given values, we get -
[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]
Also
[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]
Substituting the given values we get -
[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours
A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it
Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
10 POINTS!! SPACE QUESTION!
Answer:
The Gas Giants have more moons.
Explanation:
Mercury-0
Venus-0
Earth-1
Mars-2
Jupiter-66
Saturn-62
Uranus-27
Neptune-13
Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________
a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.
Answer:
d. the star is a member and also a part of an eclipsing binary star system.
Explanation:
If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).
The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.
Thus, option D is correct.
state newton first law of motion
Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.
got it off g lol..
Answer:
it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."
Which form of energy increases when a spring is compressed?
Answer:
When the spring compresses, elastic potential energy increases.
Answer:
the answer is b
Explanation:
elastic potential energy
The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.
Answer:
soluble soluble soluble soluble
Explanation:
solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled? *
A)60km
B)60km[W]
C)17km[W]
D) 6.6km[W]
Answer:
B)60km[W]
Explanation:
The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].
Hope that helps!
A cart of mass m is moving with negligible friction along a track with known speed v1 to the right. It collides with and sticks to a cart of mass 4m moving with known speed v2 to the right. Which of the two principles, conservation of momentum and conservation of mechanical energy, must be applied to determine the final speed of the carts, and why
Answer:
conservation of linear momentum
We were told that two objects became stuck together hence we have to use the principle of conservation of momentum to obtain the final velocities of the carts.
What is conservation of momentum ?The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.
We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.
Learn more about conservation of momentum: https://brainly.com/question/11256472
which statement can best describe the energy transformations that occur when a pendulum swings back and forth?
a. gravitational energy is converted to spring energy and back.
b. gravitational energy is converted to kinetic energy and back
c. kinetic energy is converted to spring energy and back
d. none of the above.
Answer:
I think it is C, I'm not for sure.
An object’s
✔ mass
will remain constant throughout the universe, but its
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
Answer:
mass, weight, strength of gravity increases, weight decreases
Explanation:
got it on edge
Answer:
An object’s
✔ mass
will remain constant throughout the universe, but its
✔ weight
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
✔ strength of gravity increases
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
✔ weight decreases
Explanation:
right on edge 22
The nucleus of a certain type of uranium atom contains
92 protons and 143 neutrons. What is the total charge of
the nucleus?
Answer:
charge = electrons + protons
=92+92
=184
While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Answer:
the smallest separation between two objects is 0.8067 m
Explanation:
Given the data in the question;
Altitude h = 5.75 km = 5750 m
Diameter D = 4.0 mm = 0.004 m
λ = 460 nm = 4.6 × 10⁻⁷ m
Now, Using Rayleigh criterion for Airy disks resolution.
we know that, Minimum angular separation for resolving two points is;
θ = 1.22λ / D
so we substitute
θ = (1.22 × 4.6 × 10⁻⁷) / 0.004
θ = 5.612 × 10⁻⁷ / 0.004
θ = 1.403 × 10⁻⁴ rad
so minimum separation [tex]d_{min[/tex] = θh
so we substitute
[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m
[tex]d_{min[/tex] = 0.8067 m
Therefore, the smallest separation between two objects is 0.8067 m
While visiting Planet Physics, you toss a rock straight up at 15 m/s and catch it 2.7 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet's radius every 250 min .
Part A What is the mass of Planet Physics?
Part B What is the radius of Planet Physics?
Answer:
R = 7.915 10⁶ m, M = 1.04 10³⁵ kg
Explanation:
Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations
v = v₀ - g t
the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls
v = -v₀
-v₀ = v₀ - g t
g = 2v₀ / t
g = 2 15 / 2.7
g = 11.11 m / s²
I now write the law of universal gravitation and Newton's second law
F = m a
G m M / R² = m a
a = g
g = G M / R²
Now let's work with the cruiser in orbit
F = ma
acceleration is centripetal
a = v² / r
G m M / r² = m v² / r (1)
the distance from the center of the planet is
r = R + h
r = R + R = 2R
we substitute in 1
G M / 4R² = v² / 2R
G M / 2R = v²
The modulus of the velocity in a circular orbit is
v = d / T
the distance is that of the circle
d = 2π r
v = 2π 2R / T
v = 4π R / T
G M / 2R = 16pi² R² / T²
T² = 32 pi² R³ / GM
let's write the equations
g = G M / R² (2)
T² = 32 pi² R³ / GM
we have two equations and two unknowns, so it can be solved
let's clear the most on the planet and equalize
g R² / G = 32 pi² R³ / GT²
g T² = 32 pi² R
R = g T² / 32 pi²
let's reduce the period to SI units
T = 250 min (60 s / 1 min) = 1.5 104 s
let's calculate
R = 11.11 (1.5 10⁴) ² / 32 π²
R = 7.915 10⁶ m
from equation 2 we can find the mass of the planet
M = g R² / G
M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹
M = 1.04 10³⁵ kg
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled?
Answer:
A boat travels for three hours with a... A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in still water?
Explanation:
Drag each statement to the correct location on the table.
Match the characteristics with the states of matter.
does not have
a definite shape
or volume
has definite volume
but does not have a
definite shape
has a definite shape
and volume
changes to liquid
on heating
changes to liquid
on cooling
changes to solid
on cooling
Solid
Liquid
Gas
mentum. All rights reserved.
Answer:
Solid:
- has definite shape and volume.
- change to liquid on heating.
Liquid:
- has definite volume but does not have definite shape .
- changes to solid on cooling.
Gas :
- does not have definite shape or volume.
- change to liquid on cooling
According to the Traditional Square of Opposition: If "All S are P" is true, then is "Some S are P" true or false?
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the concept of Logic.
Since in the above statement it is given that,
All S are P ==> True,
then obviously Some S are also P always, hence it is true.
Answer is True.
Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity
Answer:
C
Explanation:
Primacy means being first or important so thats not an important facial display as the others.
A car is moving with speed 30 m/s and acceleration 4 m/s2 at a given instant. (a) Using a second-degree Taylor polynomial, estimate how far the car moves in the next second.
Answer:
68 meters moved in the next seconds
Explanation:
Given
[tex]u= 30m/s[/tex]
[tex]a = 4m/s^2[/tex]
Required
Distance covered by the car in the next second
At a point in time t, the current distance is calculated as:
[tex]s(t) = ut + \frac{1}{2}at^2[/tex]
Substitute values for a and u in the above equation.
[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]
[tex]s(t) =30t + 2t^2[/tex]
Next, we generate the second degree Taylor polynomial as follows;
Calculate velocity (s'(t))
Differentiate s(t) to get velocity
[tex]s(t) =30t + 2t^2[/tex]
[tex]s'(t) =30 + 4t[/tex]
Calculate acceleration (s"(t))
Differentiate s'(t) to get acceleration
[tex]s'(t) =30 + 4t[/tex]
[tex]s"(t) =4[/tex]
When t = 0
We have:
[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]
[tex]s'(0) =30 + 4*0 = 30[/tex]
[tex]s"(0) = 4[/tex]
So, the second degree tailor series is:
[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]
To see the distance moved in the next second, we set t to 1
So, we have:
[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]
Solving s(1), s'(1) and s"(1)
We have:
[tex]s(1) =30*1 + 2*1^2 = 32[/tex]
[tex]s'(1) =30 + 4*1 = 34[/tex]
[tex]s"(1) =4[/tex]
Hence:
[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]
[tex]T_2(1) = 32 + 34 + 2[/tex]
[tex]T_2(1) = 68[/tex]
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
You need friction created by your tires and the road ____
to control your speed and direction.
Answer:
surface
Explanation:
You need friction created by your tires and the road surface
to control your speed and direction.
An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k = 450 N/m. The object is determined to have a velocity of 0.3 m/s when passing through the equilibrium.
1. Find the amplitude of the motion
2. Find the total energy of the object at any point of its motion
Answer:
1) The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object at any point of its motion is 16.892 joules.
Explanation:
1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:
[tex]U_{e} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]A[/tex] - Amplitude, in meters.
[tex]m[/tex] - Object mass, in kilograms.
[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.
If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]
[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]
[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]
[tex]A \approx 0.274\,m[/tex]
The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])
[tex]E = U_{e}[/tex]
[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]
[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]
[tex]E = 16.892\,J[/tex]
The total energy of the object at any point of its motion is 16.892 joules.