Answer:
a)
the quarterback will be moving back at speed of 0.080625 m/s
b)
the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm
Explanation:
Given the data in the question;
a)
How fast will he be moving backward just after releasing the ball?
using conservation of momentum;
m₁v₁ = m₂v₂
v₂ = m₁v₁ / m₂
where m₁ is initial mass ( 0.43 kg )
m₂ is the final mass ( 80 kg )
v₁ is the initial velocity ( 15 m/s )
v₂ is the final velocity
so we substitute
v₂ = ( 0.43 × 15 ) / 80
v₂ = 6.45 / 80
v₂ = 0.080625 m/s
Therefore, the quarterback will be moving back at speed of 0.080625 m/s
b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?
we make use of the relation between time, distance and speed;
s = d/t
d = st
where s is the speed ( 0.080625 m/s )
t is time ( 0.30 s )
so we substitute
d = 0.080625 × 0.30
d = 0.0241875 m or 2.41875 cm
Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm
Which change will always result in an increase in the gravitational force between two objects?
O increasing the masses of the objects and increasing the distance between the objects
O decreasing the masses of the objects and decreasing the distance between the objects
O increasing the masses of the objects and decreasing the distance between the objects
• decreasing the masses of the objects and increasing the distance between the objects
Answer:
increasing the masses of the objects and decreasing the distance between the objects
Explanation:
A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.
The total displacement of the dog is:
Which energy source can be found on the electromagnetic spectrum? A) sound energy B) chemical energy UV light energy D mechanical energy
Easy question just don’t understand it please help.
Please respond to this for 15 points. Please don’t put in a link.
Answer:
e. Combustion
Explanation:
In Combustion reaction, a substance reacts with oxygen from the air and resultant product is that it releases carbon dioxide and water.
Here,
2C2H6 is the substance that reacted with 7O2 (Oxygen) to release 4CO2 (Carbon Dioxide) and 6H2O (Water).
does the stirling engine follow the law of conservation energy
Answer:
Conservation of Energy: Like all things, Stirling Engines follow the conservation of energy principle (all the energy input is accounted for in the output in one form or another). ... The hot one supplies all of the energy QH, while the cold one removes energy QC (a necessary part of the cycle).
Explanation:
Answer: Yes
Explanation: All the energy input is accounted for in the output in one form or another
If a reflected ray is 55 degrees from the normal line, they what is the angle of the
incident ray from normal?
Answer:
xplanation:
Angle of reflection is measured between the incident ray and the angle which it makes with the normal at the point where incident ray strikes the mirror surface.
Further on reflection, it makes the same angle i.e. angle of reflection is equal to angle of reflection.
Hence, as angle of incidence is 55∘ angle of reflection too is 55∘ and the angle between the incident ray and the reflected ray is 55∘+55∘=110∘
The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.62c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The veolcity of Enterprise 2 relative to Enterprise 1 is 0.30c. What is the velocity of Enterprise 2
Answer:
The answer is "0.92 c"
Explanation:
[tex]v_1\ (earth) = 0.62 \ c \\\\v_2\ ( enterprise ) = -0.30[/tex]
so,
[tex]v_2 \ (earth) = 0.62 \ c - (-0.30 \ c) \\\\[/tex]
[tex]= 0.62 \ c +0.30 \ c\\\\= 0.92 \ c[/tex]
Chris used a non plane mirror to check out an box resting on a shelf. He wanted to find
the focal length of the mirror. The image of the box was located 15 cm behind the mirror
and the box was placed 19 cm from the mirror.
Chris used a non-plane mirror to check out a box resting on a shelf, the focal length of the mirror is mathematically given as
f=8.38cm
What is the focal length of the mirror?Question Parameter(s):
The image of the box was located 15 cm behind the mirror
and the box was placed 19 cm from the mirror.
Generally, the equation for the focal length is mathematically given as
1/f=1/u+1/v
Therefore
1/f=1/15+1/19
f=8.3823529cm
In conclusion, the focal length of the mirror
f=8.3823529cm
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Someone help me like please thank you
Grade 10 My smart Physics people help me with this review question please
Answer:
sorry I am not confident you the answer
What forces are used to jump over a wall?
Answer:
Potential and kinetic
Explanation:
The moment of inertia of the club head is a design consideration for a driver in golf. A larger moment of inertia about the vertical axis parallel to the club face provides more resistance to twisting of the club face for off-center hits. The mass of one club head is 200 g and its moment of inertia is 5000 g cm2 . What is the radius of gyration of this club head
Answer:
Explanation:
Moment of inertia I = M k² , where M is mass and k is radius of gyration .
Putting the given values in the equation
5000 = 200 x k²
k² = 25
k = 5 cm .
Radius of gyration is 5 cm .
what is an example of vaporization?
Answer:
just search it up you'll get ur answer
A wooden cylinder (in the form of a thin disk) of uniform density and a steel hoop are set side by side, released from rest at the same moment, and roll down an inclined plane towards a wall at the bottom. The cylinder has a larger radius than the hoop, but the hoop weighs more than the cylinder.
Required:
Who reaches the bottom first and why?
Answer:
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
Explanation:
a. Who reaches the bottom first
The kinetic energy of the objects is given by
K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²
K = 1/2mv² + 1/4mv²
K = 3mv²/4
For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop
So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²
K' = 1/2m'v'² + 1/2m'v'²
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
3mv²/4 = m'v'²
v²/v'² = 4m/3m'
v²/v'² = 4/3(m/m')
v/v' = √[4/3(m/m')]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
b. Why
Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder
K - 1/2Iω² = K₀
3/4mv² - 1/2(mr²/2)(v/r)² = K₀
3/4mv² - 1/4mv² = K₀
K₀ = 1/2mv²
For the steel hoop,
K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop
K' - 1/2I'ω'² = K₁
m'v'² - 1/2(m'r'²)(v'/r')² = K₁
m'v'² - 1/2m'v'² = K₁
K₁ = 1/2m'v'²
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
What is Kinetic energy?
The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called Kinetic energy.
The kinetic energy of the objects is given by
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]
where
m = mass of object,
v = velocity of object,
I = moment of inertia and
ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy,
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]
[tex]K = \dfrac{3mv^2}{4}[/tex]
For the steel hoop,
I' = mr'²
where
m' = mass of steel hoop and
r' = radius of steel hoop and
v' = velocity of steel hoop
So, its kinetic energy,
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]
[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]
[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]
[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
(b) Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]
where
K₀ = translational kinetic energy of wooden cylinder
[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]
[tex]K_o = \dfrac{1}{2}mv^2[/tex]
For the steel hoop,
[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]
where
K₁ = translational kinetic energy of steel hoop
[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]
[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
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Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
Which runner finished the 100 m race in the least amount of time?
Ming
Which runner stopped running for a few seconds during the race?
At what distance did Anastasia overtake Chloe in the race?
1: Ming
2: Chloe
3: 40m
A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire, in the x direction. Along 0.17 m of the length of the wire there is a magnetic field of 0.52 tesla in the y direction, due to a large magnet nearby. At other locations in the circuit, the magnetic field due to external sources is negligible. What is the magnitude of the magnetic force on the wire
Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )
given that
I = 2.6 A
[tex]B^>[/tex] = 0.17
[tex]L^>[/tex] = 0.52
so we substitute
|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )
|F[tex]_{mg}^>[/tex] | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N
At which point is there the most potential energy? At which point is there the most kinetic energy?
A. Potential energy A; Kinetic energy B
B. Potential energy B; Kinetic energy D
C. Potential energy A; Kinetic energy D
D. Potential energy C; Kinetic energy D
Answer:
The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy
Daryl ties a rope to a brick and lifts the brick straight up. The free-body
diagram below shows the brick when it is suspended above the ground.
Force 1
Force 2
What is force 1 in this diagram?
O A. Friction
OB. Tension
O C. Normal force
O D. Weight
The force 1 is tension force.
To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.
What is friction?Friction force is found between two surfaces when one is kept or moved on another surface.It is directed opposite to the direction of motion.What is tension force?When any object is hanged by an thread or rope, that object exerts a force on that rope. This force is called as tensional force.It's directed from along the rope towards the point of hanging.What is normal force?When an object is kept on a surface, the surface exerts a force on the object to oppose the weight of the object which is the normal force.It's perpendicular to the surface that an object contacts.What is weight?Weight is the gravitational force exerted by earth on that object. It's always directed towards the center of the earth.Thus, we can conclude that the correct option is (B).
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A 3.5 kg object gains 76 J of potential energy as it is lifted vertically. Find the new height of the object?
Answer:
1.72 m
Explanation:
Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height
76 = (3.5)(9.8)h
76=44.1h
h=1.72335600907 ≈1.72 m
Answer:
:r
Explanation:r
Tobnbv346468this Ishmael
find the rms speed of a sample of oxygen at 30° C and having a molar mass of 16 g/mol.
At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.
The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:
vrms = √(3kT/m)
Where:
vrms is the rms speed
k is the Boltzmann constant (1.38 x 10^-23 J/K)
T is the temperature in Kelvin
m is the molar mass of the gas in kilograms
To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:
m = 16 g/mol = 0.016 kg/mol
Substituting the values into the formula, we have:
vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))
Calculating this expression yields the rms speed of the oxygen sample:
vrms ≈ 482.34 m/s
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here is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 10.7 cm. When the cylinder is rotating at 1.65 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall
Answer:
11.5 m/s²
Explanation:
The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s
So, a = rω²
a = r(2πN)²
a = 4π²rN²
substituting the values of the variables into the equation, we have
a = 4π²rN²
a = 4π²(0.107 m)(1.65 rev/s)²
a = 4π²(0.107 m)(2.7225 rev²/s)²
a = 4π² × 0.2913075 mrev²/s)²
a = 11.5 m/s²
a lens with f = 50.0 cm is held 55.0 cm from an object. what is the image distance? (unit = cm)
Answer: 550 cm
Explanation:
Original equation: 1/f= 1/do + 1/di.
F=50.0 cm, and do=55.0.
Since we don't have di, we'll have to subtract do to the other side, making the equation: 1/f - 1/do= 1/di.
Doing the math, 1/f - 1/do is 0.0018181818
Then to get di by itself, you multiply both sides by di. Then you divide by 0.0018181818 to get di by itself. You then get: di= 1/0.0018181818
At that point, you just divide 1 by 0.0018181818, which will give you 550 cm
There could be simpler way, but that is just what I did to get the answer. Answer was right on Acellus
b. Calculate the kinetic energy of the car for group A.
Answer: Kinectic Energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity.
Explanation: If an object with a mass of 10 kg (m=10 kg) is moving at a velocity of 5 meters per second (v=5 m/s), the kinetic energy is equal to 125 Joules, or (1/2* 10 kg) * 5 m/s^2.
2.- a person weighing 70 kg travels at 2m / s. What is the value of his kinetic energy?
Answer:18 watts
Explanation:i just got this question trust me
\
Great Sand Dunes National Park in Colorado is famous for its giant sand dunes. Sand dunes are landforms that are found in deserts and on beaches. Visitors to the park can surf down the dunes on sleds or boards.
An image of sand dunes in front of a mountain and behind a body of water and grass.
Which process causes the shape of these giant dunes?
A. deposition
B. erosion
C. weathering
D. waves
Answer:
Wind deposits sand into a small mound. So the answer is Deposition
A mass of 3 kg stretches a spring 9m. The mass is acted on by an external force of 2 AND. The Mass moves in a medium that imparts a viscous force of 1 N when the speed of the mass is 4m/sec The mass is pulled down 8 cm below its equilibrium position, and then set in motion inthe upward direction with a velocity of 5 m/sec. State the initial value problem describing the motion of the mass. DO NOT SOLVE.
Answer:
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
Explanation:
In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law
Let's set a reference system with the y-axis in a vertical and positive direction upwards.
We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative
let's write Newton's second law
F_e -F -fr - W = m a
where
F_e = -kDy = - k y
fr = - b v = -b dy / dt
W = mg
we substitute for the specific case, that is, using the signs
k y -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]
In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero
k y - m g - F = 0
from this equation you can find the spring constant, y= 9m and F=2 N
It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains
k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution
The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m
A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.
The total displacement of the dog is:
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 4 inches above the equilibrium position. Find the equation of motion. (Use g
Answer:
The equation of motion is [tex]x(t)=-[/tex][tex]\frac{1}{3} cos4\sqrt{6t}[/tex]
Explanation:
Lets calculate
The weight attached to the spring is 24 pounds
Acceleration due to gravity is [tex]32ft/s^2[/tex]
Assume x , is spring stretched length is ,4 inches
Converting the length inches into feet [tex]x=\frac{4}{12} =\frac{1}{3}feet[/tex]
The weight (W=mg) is balanced by restoring force ks at equilibrium position
mg=kx
[tex]W=kx[/tex] ⇒ [tex]k=\frac{W}{x}[/tex]
The spring constant , [tex]k=\frac{24}{1/3}[/tex]
= 72
If the mass is displaced from its equilibrium position by an amount x, then the differential equation is
[tex]m\frac{d^2x}{dt} +kx=0[/tex]
[tex]\frac{3}{4} \frac{d^2x}{dt} +72x=0[/tex]
[tex]\frac{d^2x}{dt} +96x=0[/tex]
Auxiliary equation is, [tex]m^2+96=0[/tex]
[tex]m=\sqrt{-96}[/tex]
=[tex]\frac{+}{} i4\sqrt{6}[/tex]
Thus , the solution is [tex]x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}[/tex]
[tex]x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6t}[/tex]
The mass is released from the rest x'(0) = 0
[tex]=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6(0)}[/tex] =0
[tex]c_2[/tex] [tex]4\sqrt{6} =0[/tex]
[tex]c_2=0[/tex]
Therefore , [tex]x(t)=c_1[/tex] [tex]cos 4\sqrt{6t}[/tex]
Since , the mass is released from the rest from 4 inches
[tex]x(0)= -4[/tex] inches
[tex]c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}[/tex] feet
[tex]c_1=-\frac{1}{3}[/tex] feet
Therefore , the equation of motion is [tex]-\frac{1}{3} cos4\sqrt{6t}[/tex]