A simple pendulum of length 5.5 m makes 10.0 complete swings in 25 s what is the acceleration due to gravity at the location of the pendulum ?
Answer:
The acceleration due to gravity at the location of the pendulum is 34.74 m/s².
Explanation:
Given that,
The length of a simple pendulum, l = 5.5 m
It makes 10.0 complete swings in 25 s.
Frequency of pendulum,
[tex]f=\dfrac{10}{25}\\\\f=0.4\ Hz[/tex]
The time period of a simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Frequency,
[tex]f=\dfrac{1}{T}\\\\f=\dfrac{1}{2\pi \sqrt{\dfrac{l}{g}} }\\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}[/tex]
g is the acceleration due to gravity at the location where the pendulum is placed. So,
[tex]f^2=\dfrac{g}{4\pi^2l}\\\\g=f^2\times 4\pi^2l\\\\g=0.4^2\times 4\pi^2\times 5.5\\\\g=34.74\ m/s^2[/tex]
So, the acceleration due to gravity at the location of the pendulum is 34.74 m/s².
An 800-kg car moving at 20m/s takes a turn around a circle with a radius of 12.2 m.
Determine the acceleration and the net force acting upon the car.
a = 32.79 m/s²
F = = 26,232 N
Further explanationGiven
mass = 800 kg
velocity = 20 m/s
r = 12.2 m
Required
The acceleration
The net force
Solution
Circular motion :
a = v²/r
a = 20²/12.2
a = 32.79 m/s²
The net force⇒Centripetal force
F = mv²/r
F = 800 kg x 32.79 m/s²
F = 26,232 N
What is the approximate heat of water in kj/kg k?
Answer:
Specific heat (Cp) water (at 15°C/60°F): 4.187 kJ/kgK = 1.001 Btu(IT)/(lbm °F) or kcal/(kg K)
A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.70 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 88.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration
Answer:
N = 1364 N
Explanation:
given data
accelerate upward = 5.70 m/s²
mass = 88.0 kg
solution
normal force is in upward direction so, weight of the student in downward direction and acceleration is in upward direction so formula is express as
N - mg = ma ...........................1
N = m × (g+a)
put here value
N = 88.0 × (9.8 + 5.70)
N = 1364 N
A student throws a stone upward at an angle of
45° above the horizontal. Which statement best
describes the stone at the highest point that it
reaches?
(1) Its acceleration is zero.
(2) Its acceleration is a minimum, but not zero.
(3) Its gravitational potential energy is a
minimum
(4) Its kinetic energy is a minimum
Answer:
(4) Its kinetic energy is a minimum.
Explanation:
Stone experiments a parabolic motion, which is a combination of horizontal motion at constant speed and vertical uniform accelerated motion due to gravity, where effects from air viscosity and Earth's rotation are neglected. Meaning that stone represents a conservative system.
When stone reachest highest point, horizontal velocity remains unchanged and vertical velocity is zero. Acceleration remains constant and different of zero. Hence, gravitational potential energy is a maximum and kinetic energy is a minimum.
Correct answer is: (4) Its kinetic energy is a minimum.
The highest point, that stone reaches its kinetic energy is a minimum. therefore the option 4 is correct.
Projectile motion -The motion of an object, thrown (projected) into the air is known as projectile motion.
(1) Its acceleration is zero-
When an object is at its highest point the acceleration will be equal to the gravitational force (9.8 m/sec squared). If we take air resistance into the account it will be slightly greater than the 9.8 meters per second squared but not equal to zero in any case. Hence, statement 1 is incorrect.
(2) Its acceleration is a minimum, but not zero-
At the highest point, the object will be at the one place where air resistance does not affect the object, and thus acceleration is exactly equal to the acceleration due to gravity and at this position, it will be the maximum. Hence, statement 2 is incorrect.
(3) Its gravitational potential energy is a minimum-
At the highest point, the object will stop for the moment and have zero velocity. Thus it will have zero kinetic energy. Therefore total energy will have in the form of gravitational potential energy and which is maximum at this point. Hence, statement 3 is incorrect.
(4) Its kinetic energy is a minimum-
At the highest point the object will stop for the moment and have zero velocity. Thus it will have zero kinetic energy. Hence, statement 4 is correct.
At the highest point, that stone reaches its kinetic energy is a minimum. therefore the option 4 is correct.
For more about the projectile motion, follow the link below-
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what is air resistance means explain it with free falling body
Emma is working in a shoe test lab measuring the coefficient of friction for tennis shoes on a variety of surfaces. The shoes are pushed against the surface with a force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine has to pull with a force of only 200 N to keep the material moving.
a. What is the coefficient of static friction between the shoe and the material?
b. What is the coefficient of dynamic friction between the shoe and the material?
c. Draw a Free Body Diagram for the above.
Answer:
Explanation:
Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .
a )
Given,
Normal force N = 400 N
Force of friction = 300 N
μ = coefficient of static friction = ?
Putting the values ,
300 = 400 μ
μ = .75
b )
Normal force N = 400 N
Force of friction = 200 N
μ = coefficient of kinetic friction = ?
Putting the values ,
200 = 400 μ
μ = .50
c ) see attached file .
Arrange the objects in order from greatst to least of potential energy assume that gravity is constant
Answer:
Water > Box of books > Stone > Ball
Explanation:
We'll begin by calculating the potential energy of each object. This can be obtained as follow:
For stone:
Mass (m) = 15 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 m
Potential energy (PE) =?
PE = mgh
PE = 15 × 10 × 3
PE = 450 J
For water:
Mass (m) = 10 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 9 m
Potential energy (PE) =?
PE = mgh
PE = 10 × 10 × 9
PE = 900 J
For ball:
Mass (m) = 1 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 20 m
Potential energy (PE) =?
PE = mgh
PE = 1 × 10 × 20
PE = 200 J
For box of books:
Mass (m) = 25 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 2 m
Potential energy (PE) =?
PE = mgh
PE = 25 × 10 × 2
PE = 500 J
Summary:
Object >>>>>>>> Potential energy
Stone >>>>>>>>> 450 J
Water >>>>>>>>> 900 J
Ball >>>>>>>>>>> 200 J
Box of books >>> 500 J
Arranging from greatest to least, we have:
Object >>>>>>>> Potential energy
Water >>>>>>>>> 900 J
Box of books >>> 500 J
Stone >>>>>>>>> 450 J
Ball >>>>>>>>>>> 200 J
Water > Box of books > Stone > Ball
Determine mass flow rate and velocity of efflux from circular hole of 0.1 diameter at the bottom of water tank at this instant .
The tank is open to atmosphere and H=4m
Answer:
Mixed in a smoothie it like it licked
Explanation:
Determine the magnitude of the gravitational force (in nano-Newton's, i.e. 10^-9N) between a 61.6 kg girl and a 71.2 kg boy standing 95 m apart from one another.
Answer:
3.24×10¯² nN
Explanation:
From the question given above, the following data were obtained:
Mass of girl (M₁) = 61.6 kg
Mass of boy (M₂) = 71.2 kg
Distance apart (r) = 95 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force (F) =?
The force of attraction between the girl and the boy can be obtained as follow:
F = GM₁M₂ /r²
F = 6.67×10¯¹¹ × 61.6 × 71.2 / 95²
F = 6.67×10¯¹¹ × 61.6 × 71.2 / 9025
F = 3.24×10¯¹¹ N
Finally, we shall convert 3.24×10¯¹¹ N to nN. This can be obtained as follow:
1 N = 10⁹ nN
Therefore,
3.24×10¯¹¹ N = 3.24×10¯¹¹ N × 10⁹ nN / 1 N
3.24×10¯¹¹ N = 3.24×10¯² nN
Thus, the force attraction between the girl and the boy is 3.24×10¯² nN
The "problem of perception" is best characterized as?
Answer:
making sense of a 3-d world from 2-d data
Explanation:
___is found in milk, made from glucose and galactose. *
1.Maltose
2.Xylose
3.Lactose
4.Sucrose
Answer:
lactose
Explanation:
A 20- kg and a 50-kg ball rest at the top of a ramp. Which statement best describes the potential energy of the balls?
Answer:The 50-kg has the greater potential energy because it has greater mass
Explanation:
In ionic compounds which group from the periodic table usually provide anion?
A sealed container with a lid of area 0.004 m^2 is filled with an ideal gas. The container and gas are allowed to reach thermal equilibrium with the surrounding air. If a 2000 N block is needed to keep the lid from being pushed off the container, what is the absolute pressure inside the container (the pressure compared to vacuum)
Answer:
6 * 10^5 N/m²
Explanation:
Given that :
Area of lid = 0.004m²
Force of block needed to keep the lid from being pushed off the container = 2000 N
Absolute Pressure = atmospheric pressure + force / Area
Force / Area = 2000 N / 0.004 m² = 500,000 = 5 * 10^5
Atmospheric pressure = 1.01325 * 10^5 N/m²
Absolute Pressure = (1.01325 * 10^5) + (5 * 10^5)
Absolute Pressure = 6.01325 * 10^5
= 6 * 10^5 N/m²
2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is placed in a centrifuge (r = 25.0 m) and spun at a constant angular velocity of 10.0 rpm (revolutions per minute).
a. Find the linear velocity of the centrifuge in m/s. Show your work
b. Find the magnitude and direction of the centripetal acceleration when he is spinning at this constant velocity.
c. How many g’s is the astronaut experiencing? (at constant velocity)
d. Find the linear deceleration and torque required to bring the centrifuge (5000.0 kg) to a stop over a 5 minute time period.
Answer:
a) v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c) a = 2.8 g and
d) a = - 8.73 10⁻² m / s², τ = 1.09 10⁴ N m
Explanation:
a) For this exercise we can use the relationships between rotational and linear motion
v = w r
let's reduce the magnitudes to the SI system
w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s
r = 25.0 m
let's calculate
v = 1.047 25.0
v = 26.2 m / s
b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is
a = v² / r
a = 26.2²/25
a = 27.4 m / s²
c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity
a / g = 27.4 / 9.8
a / g = 2.8
a = 2.8 g
d) let's find the deceleration and torque to stop the centripette in 5 min
t = 5 min (60 s / 1min) = 300 s
let's use the rotational kinematics relations
w = w₀ + α t
initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0
α = - w₀ / t
α = - 1,047 / 300
α = -3.49 10⁻³ rad / s²
angular and linear are related
a = α r
a = -3.49 10⁻³ 25
a = - 8.73 10⁻² m / s²
the negative sign indicates that the acceleration is stopping the movement
torque is
τ = F r
The force can be found with Newton's second law
F = m a
we substitute
τ = m a r
τ = 5000.0 8.73 10⁻² 25
τ = 1.09 10⁴ N m
How is light energy different from both sound and heat energy?
A.
Sound energy is a wave.
B.
Light energy can exist independently of matter.
C.
Heat energy can radiate from an object.
D.
Light is more energetic when its frequency is higher.
Answer:
B. Light energy can exist independently of matter.
Explanation:
In science, matter can be defined as anything that has mass and occupies space. Any physical object that is found on earth is typically composed of matter. Matter are known to be made up of atoms and as a result has the property of existing in states.
Generally, matter exists in three (3) distinct or classical phases and these are;
I. Solid.
II. Liquid.
III. Gas.
Light energy is different from both sound and heat energy because light energy can exist independently of matter.
Generally, we know that light energy is a type of electromagnetic waves and as such do not require a medium of propagation for it to travel through a vacuum of space where no particles exist. For example, light energy from distant galaxy of stars and the sun.
However, both sound and heat energy are mechanical waves which are highly dependent on matter for their propagation and transmission.
a ball has a mass of 140g if it thrown with a velocity of 450m/s what is its kenetic energy?
Kinetic energy = mv²
Therfore kinetic energy =14175 joule
2. What is the cheetah's speed for the first four seconds. She
Explanation:
Cheetahs can go from 0 to 60 miles per hour in just 3.4 seconds and reach a top speed of 70 miles per hour. While they are the fastest land animal in the world, they can only maintain their speed for only 20 to 30 seconds.
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 49.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 49.0 mi/h. At the same time, the cyclist speeds up from rest to 21.0 mi/h with constant acceleration 14.00 mi/h/s and thereafter moves with a constant speed of 21.0 mi/h.
Required:
For what time interval (in s) after the light turned green is the bicycle ahead of your car?
Answer:
t = 4.34 s
Explanation:
To get this, we need to calculate the time of each part for both vehicles.
In the case of the bicycle, we can calculate the time duration with it's acceleration:
a = Vf - Vo/t ------> t = Vf - Vo / a
But Vo = 0 so time:
t = 21 / 14 = 1.5 s
Doing the same with the car:
t = 49 / 8 = 6.125 s
With these values of time, we can calculate the distance covered by both vehicles during acceleration:
X = Vo*t + at²/2
X = at² / 2
With the bicycle:
X = 14 * (1.5)² / 2 = 15.75 mi
With the car:
X = 8 * (1.5)² / 2 = 9 mi
And now, we can also get the speed of the car, after the time of 1.5 s has passed.
V = 8 * 1.5 = 12 mi/h
Now we can actually write an equation with both data of the vehicles in function of the distance. We are going to say that "t" would be the time taken by both vehicles, to meet each other.
Distance covered by the bicycle = distance covered by car
Distance covered by the bicycle would be:
15.75 mi + 21t
Distance covered by car:
9 + 12t + (8t²/2)
Equalling both expressions:
15.75 + 21t = 9 + 12t + 4t²
4t² - 9t - 6.75 = 0
Solving for t, using quadratic expressions we have:
t = 9 ± √(9)² + 4*4*6.75 / 2*4
t = 9 ±√189 / 8
t = 9 ± 13.75 / 8
t₁ = 2.84 s
t₂ = -0.594 s
We are taking the positive time.
Then, the time where both vehicles meet, which is the same time interval when the bicycle is ahead of the car will be:
t = 2.84 + 1.5
t = 4.34 sHope this helps
What three factors determine the amount of potential energy in a object are ______,______,and ______.
Answer:
It should be Mass, Gravity and Height
Explanation:
An object is attached to the lower end of a 100-coil spring that is hanging from the ceiling. The spring stretches by 0.150 m. The spring is then cut into two identical springs of 50 coils each. Each spring is attached between the ceiling and the object. By how much does each spring stretch?
Answer:
0.125m
Explanation:
Given that:
Length of spring = 100m
Extension, e = 0.250
Recall :
F = ke - - - - (1)
Therefore, if spring is cut into 2 halves :
Then,
k = k1 + K2 ; since k1 and K2 are of equal length
Extension = e'
k = 2k
F = ke'
Then,
F = 2ke' - - - - - (2)
Equate (1) and (2)
ke = 2ke'
e = e'
0.250 = 2e'
0.250 / 2 = e'
e' = 0.125m
Calculate the density of a substance that has mass 10g and volume 2mL
You're supposed to divide the mass by the volume, which is going to equal to 5
Grace drives her car 168 km in 2 hours. What is her average speed in kilometers per hour?
Answer:
84kliometers
Explanation:
divide one hundred and sixty eight kilo meters by two hours
Which type of force describes the force applied to an object?
(10 Points)
Normal Force
Force of Gravity
Applied Force
Force of Friction
Tension Force
Plssss help!!!!
Answer: Force of Gravity
Explanation:
Because if a force is applied to an object then the force of gravity will move the object.
Hope this helps you!
in a football game, the kicker kicks a football a horizontal distance of 43 yards if the ball lands 3.9 seconds later, what is the balls horizontal velocity
Answer:
10s
Explanation:
Horizontal velocity is the velocity of an object in an horizontal direction
The ball's horizontal velocity is approximately 33.078 ft./s
Reason:
The known parameter are;
The horizontal distance the footballer kicks the ball, d = 43 yards
The time after which the ball lands, Δt = 3.9 seconds
Required:
To find the velocity of the ball
Solution:
[tex]Velocity = \dfrac{Distance}{Time} = \dfrac{d}{\Delta t}[/tex]
Therefore;
[tex]Horizontal \ velocity \ of \ the \ ball, \ v_x= \dfrac{43 \ yard}{3.9 \ seconds} \approx 11.026 \ yd/s[/tex]
The ball's horizontal velocity, vₓ ≈ 11.026 yd/s
1 yard = 3 feet
[tex]11.026 \ yard = 11.026 \ yard \times \dfrac{3 \ feet}{yard} = 22.078 \ feet[/tex]
The ball's horizontal velocity, vₓ ≈ 33.078 ft./s
Learn more about horizontal velocity here:
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A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the coil makes an angle of 30∘ with respect to the magnetic field.
The flux through the coil is
Answer:
1.5 * 10^-2 Tm^2
Explanation:
Electric Flux = B.A cos(theta)
B = 0.055 T
A = 0.32 m^2
theta = 30
Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2
A 0.30-m radius car tire rotates how many rad after starting from rest and accelerating at a constant 3.0 rad sa
over a 5.0-s interval?
Answer:
The angular displacement is 37.5 radian.
Explanation:
Given that,
The radius of the car, r = 0.3 m
The acceleration of the car, [tex]\alpha =3\ rad/s^2[/tex]
The initial speed of the car, [tex]\omega_i=0[/tex]
Time, t = 5 s
The angular displacement can be calculated using second equation of motion i.e.
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\\theta=\dfrac{1}{2}\alpha t^2\\\\\theta=\dfrac{1}{2}\times 3\times (5)^2\\\\\theta=37.5\ rad[/tex]
So, it will make 37.5 radians.
A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of charge distribution is inside the surface? a positively charged plane parallel to the end faces of the cylinder a positive line charge situated on and parallel to the axis of the cylinder a collection of positive point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a collection of negative point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a negatively charged plane parallel to the end faces of the cylinder (b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface? How is the electric flux related to the electric field vector and the normal to the surface? What is the orientation of the electric field relative to the curved surface? N · m2/C (c) What is the net charge inside the cylinder?
Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,[tex]\phi=-2 EA=-2E(\pi r^2)[/tex]
[tex]\phi=-2\times 300\times (3.14\times (0.66)^2)[/tex]
[tex]\phi=-820.67 Nm^2/C[/tex]
c.
Net charge,[tex]Q=\epsilon_0\times \phi[/tex]
Where
[tex]\epsilon_0=8.85\times 10^{-12}[/tex]
[tex]Q=-820.67\times 8.85\times 10^{-12}[/tex]
[tex]Q=-7.26\times 10^{-9} C[/tex]
[tex]Q=-7.26nC[/tex]
Where [tex]1nC=10^{-9}C[/tex]
A 3.50 kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of T(t) = (36.0 N/s)t. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity.
Required:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
b. What is the maximum distance that the box descends below its initial position?
c. At what value of t does the box return to its initial position?
Answer:
a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s
Explanation:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
We write the equation of the forces acting on the mass.
So, T - mg = ma where T = tension in vertical rope = (36.0 N/s)t, m = mass of box = 3.50 kg, g = acceleration due to gravity = 9.8 m/s² and a = acceleration of box = dv/dt where v = velocity of box and t = time.
So, T - mg = ma
T/m - g = a
dv/dt = T/m - g
dv/dt = (36.0 N/s)t/3.50 kg - 9.8 m/s²
dv/dt = (10.3 m/s²)t - 9.8 m/s²
dv = [(10.3 m/s²)t - 9.8 m/s²]dt
Integrating, we have
∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt
∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt
v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C
v = (5.15 m/s³)t² - (9.8 m/s²)t + C
when t = 0, v = 0 (since at t = 0, box is at rest)
So,
0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C
0 = 0 + 0 + C
C = 0
So, v = (5.15 m/s³)t² - (9.8 m/s²)t
i. What is the velocity of the box at t = 1.00 s,
v = (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)
v = 5.15 m/s - 9.8 m/s
v = -4.65 m/s
ii. What is the velocity of the box at t = 3.00 s,
v = (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)
v = 15.45 m/s - 29.4 m/s
v = -13.95 m/s
b. What is the maximum distance that the box descends below its initial position?
Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t
dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt
Integrating, we have
∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt
∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt
∫dy = ∫(5.15 m/s³)t³/3dt - ∫(9.8 m/s²)t²/2dt
y = (1.72 m/s³)t³ - (4.9 m/s²)t² + C'
when t = 0, y = 0.
So,
0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'
0 = 0 + 0 + C'
C' = 0
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
The maximum distance is obtained at the time when v = dy/dt = 0.
So,
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0
(5.15 m/s³)t² - (9.8 m/s²)t = 0
t[(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or (5.15 m/s³)t = (9.8 m/s²)
t = 0 or t = (9.8 m/s²)/(5.15 m/s³)
t = 0 or t = 1.9 s
Substituting t = 1.9 s into y, we have
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²
y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)
y = 11.798 m - 17.689 m
y = -5.891 m
y ≅ - 5.89 m
So, the maximum distance that the box descends below its initial position is 5.89 m
c. At what value of t does the box return to its initial position?
The box returns to its original position when y = 0. So
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
0 = (1.72 m/s³)t³ - (4.9 m/s²)t²
(1.72 m/s³)t³ - (4.9 m/s²)t² = 0
t²[(1.72 m/s³)t - (4.9 m/s²)] = 0
t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0
t = √0 or (1.72 m/s³)t = (4.9 m/s²)
t = 0 or t = (4.9 m/s²)/(1.72 m/s³)
t = 0 or t = 2.85 s
So, the box returns to its original position when t = 2.85 s