Answer:
Di ko alam Basta
mag sikap ka na lng
A bike and rider of total mass 75.0 kg moving at a velocity of 30 m/s to the right
collides with a skater of mass 60,0 kg moking at a velocity of 6.0 m/s to the left.
After they collide, the skater has a velocity of O m/s. What is the final velocity of the
bike and rider?
O A. O m/s
O B. 1.8 m/s to the left
O C. 3.0 m/s to left
O D. 1.8 m/s to the right
Answer:
[tex](75.0 \times 30) + (60.0 \times 6.0) = (75.0 \times V) + (60.0 \times 0) \\ 2250 + 360 = 75V \\ 75V = 2610 \\ V = 34.8 \: m {s}^{ - 1} [/tex]
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)
Answer:
144 meters
Explanation:
the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t
S (height cliff)=24m/s×6s=144
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
The question is incomplete, the complete question is;
In addition to absorption of a photon, energy can be transferred to an atom by collision. Consider a hydrogen atom in its ground state. Incident on the atom are electrons having a kinetic energies of 10.5 eV. What is a possible result?
A) The atom moves to a state of lower energy
B) The atom is ionized
C) One of the electrons leaves the atom
D) The atom can be excited to a higher energy state
Answer:
The atom can be excited to a higher energy state
Explanation:
According to the Bohr model of the atom, electrons in an atom can be excited from a lower to a higher energy level when energy is absorbed by the atom.
If electrons having an energy of 10.5ev are incident on a hydrogen atom, this energy is transferred to the atom by collision. Since the energy transferred is less than the ionization energy of hydrogen atom in its ground state(13.6ev), the atom is not ionized.
Rather, the atom is excited from ground state to a higher energy level.
Question 10 of 10
What does a resistor in an electrical circuit do?
A. It attracts electrons, causing them to move faster.
B. It allows only a reduced number of electrons to flow through it.
C. It opposes the flow of electrons.
D. It repels electrons, causing them to return to the battery.
A resistor in an electrical circuit opposes the flow of electrons by providing resistance to the current. The correct option is C
What is resistor ?A resistor is a passive electrical device that controls or restricts how much electrical current can flow across a circuit in an electronic device.
High electrical resistance materials, such carbon, metal film, or wirewound, are used to make resistors. Ohms (Ω) are used to measure a resistor's resistance. A resistor's resistance determines how much it will fight against the flow of electrical electricity.
Therefore, A resistor's resistance is the degree of opposition it offers. The more resistance present, the more challenging it is for electrons to move across the circuit.
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Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,
Answer:
A
Explanation:
If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.
Answer: C.
Explanation: plato users
Help help help HELP AAAAA
It's nighttime, and you've dropped your goggles into a 3m deep swimming pool. If you hold a laser pointer 1.0m abovve the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0m from the edge. How far are the goggles from the edge of the pool?
Answer:
x_total = 3.07 m
Explanation:
Let's analyze the situation presented, in this case we have the data to find the angle of incidence and with the law of refraction we can find the angle of refraction.
Let's start looking for the angle with which the laser pointer reaches the water, let's use trigonometry
tan θ₁ = 1.0 /2.0
θ₁ = tan⁻¹ 0.50
θ₁ = 26.57º
now let's use the law of refraction to find the angle of refraction (in water)
n₁ sin θ₁ = n₂ sin θ₂
the refractive index of air is n₁ = 1 and that of water n₂ = 1.33
sin θ₂ = [tex]\frac{n_1}{n_2} \ sin \theta_1[/tex]
sin θ₂ = [tex]\frac{1}{1.33} \ sin \ 26.57[/tex]
θ₂ = sin⁻¹ 0.3363
θ₂ = 19.65º
now we can find the distance from the entry point to the water to the lenses
tan θ₂ = x₂ / 3
x₂ = 3 tan θ₂
x₂ = 3 tan 19.65
x₂ = 1.07 m
the total distance from the edge of the pool is
x_total = 2 + x₂
x_total = 2 + 1.07
x_total = 3.07 m
small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becomes hot. The radiator is made of copper, and the coolant has a coefficient of volume expansion of 410 x 10-6 [1/ oC]. If the radiator is filled to its 15 [quart] capacity when the engine is cold at 6.0 [oC], how much overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 [oC]
Answer:
0.53 quart
Explanation:
The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C
Since, ΔV = VγΔθ
substituting the values of the variables into the equation, we have
ΔV = VγΔθ
ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C
ΔV = 528900 × 10⁻⁶ quart
ΔV = 0.528900 quart
ΔV ≅ 0.53 quart
Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.
an object that has momentum must have
Answer:
Mechanical energy
Explanation:
If an object does not have any momentum, then it doesn't have mechicanl energy.
Which describes farsightedness? O Distant objects are blurry. O Concave lenses can correct it. O Objects appear larger when wearing corrective glasses. O Corrective glasses do not change apparent the size of objects.
Answer:
O Distant objects are blurry. describes farsightedness.
Explanation:
Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry. The degree of your farsightedness influences your focusing ability.Farsightedness (hyperopia) is a common vision condition in which you can see distant objects clearly, but objects nearby may be blurry.
For each value of the principal quantum number n, what are the possible values of the electron spin quantum number m_s?
a. 0
b. -3/2
c. +3/2
d. +1/2
e. -1/2
Answer:
d) and e)
Explanation:
For any value of the principal quantum n, the only possible values of the electron spin quantum number m_s are + 1/2 and -1/2.Since this number is related with the angular momentum of the electron, it has a magnitude (1/2) and an orientation given by the sign (+ or -)So, the only right answers are d) and e).When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.
Answer: When it comes to the movement of air, friction is greater near the ground surface.
Explanation:
A resistance in motion observed by an object while on another object is called friction.
For example, a vehicle moving on road will have friction between its tires and the road.
Friction is more near the ground surface rather than away from the ground surface.
Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.
Ethyl alcohol is :
a. None of the above
b. Semi polar solvent
c. Polar solvent
d. Non-Polar solvant
Answer:
D. Non- polar solvant
Explanation:
l think that's it
Answer:
I think the answer is D polar solvent
which two changes would make this reaction favor the formation of products
Answer:
C. ...
D. ...
Given.
Required.
Solution.
As the temperature decreases, the equilibrium will shift towards the exothermic reaction, so the reaction shifts to the right towards SO₃( products-favored)
And increasing the pressure, then the reaction shifts to the right SO₃( products-favored)⇒the number of coefficients is greater.
Answer:
decrease the temperature
and
raising the pressure
Explanation:
In an experiment to measure the temperature of a Bunsen burner flame, a 250 g piece of iron is held in the flame for several minutes until it reaches the same temperature as the flame . The hot metal is then quickly transferred to 285 g of water contained in a 40.0 g copper calorimeter at 15.0 oC. The final temperature of the copper and water is 80.0 oC.
Using your answer from determine the temperature of the Bunsen flame.
Answer:
wait
Explanation:
Una pelota de basket es soltada desde 2.5 m de altura y rebota con una velocidad igual a 3/4 partes de la velocidad que llego. ¿ a qué altura alcanza la bola en el rebote ? ¿ cuánto tiempo transcurre desde que rebota ?
Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J at point A. The particle subsequently passes through point B which has an electric potential of -2.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.
Answer:
16.9000000000000001 J
Explanation:
From the given information:
Let the initial kinetic energy from point A be [tex]K_A[/tex] = 1.9000000000000001 J
and the final kinetic energy from point B be [tex]K_B[/tex] = ???
The charge particle Q = 6 mC = 6 × 10⁻³ C
The change in the electric potential from point B to A;
i.e. V_B - V_A = -2.5 × 10³ V
According to the work-energy theorem:
-Q × ΔV = ΔK
[tex]-Q \times ( V_B - V_A) = (K_B - K_A)[/tex]
[tex]-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]15 = (K_B - 1.9000000000000001 \ J)[/tex]
[tex]K_B = 15+ 1.9000000000000001 \ J[/tex]
[tex]\mathbf{K_B =1 6.9000000000000001 \ J}[/tex]
12. What type of circuit is the diagram below?
series circuit
parallel circuit
Answer:
parallel circuit
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;
I. Series circuit
II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.
Hence, the type of circuit that the above diagram above represents is a parallel circuit.
Answer:
parallel circuit
Explanation:
I got it right on my exam
A circular solenoid has 100 loops per centimeter, a radius of 4.53 cm, a length of 8.88 cm, and carries a current of 500 mA. It's inductance is:____________
a) 7.75 mH
b) 71.9 mH
c) 18.0 mH
d) 0.912 mH
Answer:
The inductance of the solenoid is 71.98 mH
Explanation:
Given;
number of loops per length of the solenoid, n = 100 loops/cm
radius of the solenoid, r = 4.53 cm
length of the solenoid, L = 8.88 cm
current carried by the solenoid, I = 500 mA
The inductance of the solenoid is calculated as;
[tex]L = \frac{N^2\times \mu_0 \times A}{l}[/tex]
where;
A is the area of the coil
[tex]A =\pi r^2 = \pi (0.0453)2 = 0.00645 \ m^2[/tex]
N is the number of turns, = 100 x 8.88 = 888 loops
l is the length = 0.0888 m
[tex]L = \frac{N^2\times \mu_0 \times A}{l}\\\\L = \frac{(888)^2\times \(4\pi \times 10^{-7}) \times (0.00645)}{0.0888}\\\\L = 0.07198 \ H\\\\L = 71.98 \ mH[/tex]
Therefore, the inductance of the solenoid is 71.98 mH
An ideal heat engine operates between two
temperatures 367.7 K and 865.1 K. What is
the efficiency of the engine?
0.787
0.42504
None
0.213
0.5750
Answer:
As it is the most efficient heat engine, it's efficiency is [T1 - T2]/T1. It can be measured for every Carnot cycle. From the formula and diagram, we can understand that the efficiency of an ideal heat engine also depends on the difference between the hot and cold reservoirs.
Explanation:
What is the potential difference per unit length between two infinitely long concentric cylindrical shells with inner radius 1.5 cm and outer radius 5.6 cm, if the charge on the inner and outer cylinders are 7.0 nC and -7.0 nC, respectively
Answer:
165.8 V/m
Explanation:
The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)
Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)
V = Q/C
V = Q ÷ 2πε₀L/㏑(b/a)
V = Q㏑(b/a)/2πε₀L
So, the potential difference per unit length V' is
V' = V/L = Q㏑(b/a)/2πε₀
Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have
V' = Q㏑(b/a)/2πε₀
V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)
V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)
V' = 0.16575 × 10³ V/m
V' = 165.75 V/m
V' ≅ 165.8 V/m
A particle of unit mass moves so that displacement after t seconds is given by x = 2 cos (t - 2). Find the acceleration and kinetic energy at the end of 3 seconds. (K.E = (1/2) m v²)
Answer:
a₃ = -1.08 m/s², K = 1.42 J
Explanation:
The particle is in a periodic motion, so the general expression is
x = A cos (wt + Ф)
let's compare the terms with the expression they give us
x = 2 cos (t - 2)
the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.
to find the acceleration we use its definition
v = dx / dt
a = dv / dt
a = [tex]\frac{ d^2x}{dt^2}[/tex]
let's perform the derivative
v = - A w sin (wt + Ф)
a = - A w² cos wt + Ф)
substituting the values
a = - 2 1² cos (t-2)
for t = 3 s
a₃ = 2 cos (3-2)
remember angles are in radians
a₃ = -1.08 m/s²
To calculate kinetic energy, let's find the velocity for t = 3 s
v = - 2 sin (t-2)
v = -2 sin (3-2)
v = - 1.683 m / s
body mass is m = 1 kg
we calculate
K = ½ m v²
K = ½ 1 (-1.683) ²
K = 1.42 J
An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60 degrees above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff
Answer:
The height of the cliff is 27.02 m
Explanation:
Given;
height above the ground from which the arrow is shot, h₁ = 1.5 m
initial velocity of the projectile, v = 30 m/s
angle of projection, θ = 60⁰
time taken to reach top of the cliff, t = 4.0 s
The vertical component of the velocity is calculated as;
[tex]v_y = v \times sin(\theta)\\\\v_y = 30 \times sin(60)\\\\v_y = 25.98 \ m/s[/tex]
The height attained by the projectile at the given time is calculated as;
[tex]h_2= v_y t - \frac{1}{2}gt^2\\\\h_2 = 25.98\times 4 \ - \ \frac{1}{2} \times 9.8\times 4^2\\\\h_2= 103.92 \ - \ 78.4 \\\\h_2 = 25.52 \ m[/tex]
The height of the cliff is calculated as;
H = h₁ + h₂
H = 1.5 m + 25.52 m
H = 27.02 m
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
I really don’t know the answer for this
The time-average power carried by a UPEMW propagating in vacuum is 0.05 W/m2. i) What is the amplitude value of the electric field and the amplitude value of the magnetic field in the wave
Answer:
The correct solution is "11.51 mA".
Explanation:
Given:
Time average power,
[tex]P_{avg}=0.05 \ W/m^2[/tex]
n = 377
As we now,
⇒ [tex]P_{avg}=\frac{E_0^2}{n}[/tex]
or,
⇒ [tex]E_0^2=0.05\times 377[/tex]
⇒ [tex]=4.341 \ V[/tex]
hence,
⇒ [tex]H_0=\frac{E_0}{n}[/tex]
By putting the values, we get
[tex]=\frac{4.341}{377}[/tex]
[tex]=11.51 \ mA[/tex]
what is electricity ?
Answer:
it is energy resulting in charged particles
Electricity a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current
A 55 kg pole vaulter falls from rest from a height of 5.4 m onto a foam rubber pad. The pole vaulter comes to rest 0.24 s after landing on the pad.
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision.
Answer:
a) 10.3 m/s
b) 566 N
Explanation:
[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]
[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]
The athelete velocity will be 10.3 and constant force 566 N.
What is velocity?The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).
Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.
The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.
Therefore, The athelete velocity will be 10.3 and constant force 566 N.
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A 1540-kg truck has a wheel base of 3.13 m (this is the distance between the front and rear axles). The center of mass of the truck is 1.09 m behind the front axle.a) What is the force exerted by the ground on each of the front wheels?b) What is the force exerted by the ground on each of the back wheels? Hint: Remember that the truck has four wheels.
a. The force exerted by the ground on each of the front wheels is 4918.16 Newton.
b. The force exerted by the ground on each of the back wheels is 2627.84 Newton.
Given the following data:
Mass of truck = 1540 kgDistance between the front and rear axles = 3.13 meters.Center of mass of the truck = 1.09 meters.a. To determine the force exerted by the ground on each of the front wheels:
First of all, we would take moment about the rear wheels.
[tex]F(3.13) - 1540(9.8) \times (3.13 - 1.09) = 0\\\\3.13F - 15092 \times 2.04 =0\\\\3.13F -30787.68=0\\\\F=\frac{30787.68}{3.13}[/tex]
Force, F = 9836.32 Newton
For each front wheel:
[tex]Force = \frac{9836.32}{2}[/tex]
Force = 4918.16 Newton.
b. To determine the force exerted by the ground on each of the back wheels:
We would determine the sum of the vertical forces acting on the wheels.
[tex]9836.32 + B - 1540(9.8) = 0\\\\9836.32 + B - 15092 = 0\\\\B=15092-9836.32[/tex]
B = 5255.68 Newton.
For each back wheel:
[tex]Force = \frac{5255.68}{2}[/tex]
Force = 2627.84 Newton.
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A mass is suspended on a spring. The spring is compressed so that the mass is located 5 cm above its rest position. The mass is released at time t= 0 and allowed to oscillate. It is observed that the mass reaches its lowest point 1/2s after it is released.
Required:
Find an equation that describes the motion of the mass.
Answer:
y = 5 cos 2πt
Explanation:
We will use the formula for simple harmonic motion curve where;
y = a cos ωt
Where;
a is amplitude
t is period
ω is angular frequency with the formula; ω = 2π/t
We are told that when the spring is compressed, the mass is located 5 cm above its rest position.
Thus;
a = 5 cm
it's highest point is 5 cm, but we are told that after 1/2 second of being released, it reaches its lowest point.
Since highest point is 5, then lowest point will be -5.
The difference in time between the highest and lowest point is ½ s. Which is half of the period.
Thus;
t/2 = ½
Thus, t = 1 s
Now, we know that;
t = 1/f = 2π/ω
Since t = 1, then 1 = 1/f
f = 1
Thus;
2π/ω = 1
Thus, ω = 2π
Thus, the equation is;
y = 5 cos 2πt
The equation that describes the motion of the mass is y = 5 cos 2πt.
The given parameters;
maximum displacement of the spring, A = 5 cmtime taken for the mass to reach the lowest point (half period), t = 0.5 sThe general equation of the wave is given as;
[tex]y = A\ cos\ \omega t[/tex]
where;
A is the amplitude of the vibrationω is the angular speed of massThe angular speed of the mass is calculated as;
[tex]\omega = 2\pi f\\\\[/tex]
The period of the oscillation is calculated as;
[tex]T = 2t \\\\T = 2(0.5 s) = 1 \ s[/tex]
The frequency of the wave is calculated as;
[tex]f = \frac{1}{T} \\\\f = \frac{1}{1} \\\\f = 1\ Hz[/tex]
The equation that describes the motion of the mass is calculated as;
[tex]y = A\ cos \ \omega t\\\\y = A\ cos \ 2\pi ft\\\\y = 5\ cos \ 2\pi (1) t\\\\y = 5 \ cos \ 2\pi t[/tex]
Thus, the equation that describes the motion of the mass is y = 5 cos 2πt.
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