The concentration of all species in a 0.100 M H₃PO₄ solution is as follows: [H₃PO₄] = 0.100 M, [H₂PO₄⁻] = 0.045 M, [HPO₄²⁻] = 0.0049 M, and [PO₄³⁻] = 1.0 x 10^-7 M.
Phosphoric acid, also known as orthophosphoric acid, is a triprotic acid with the chemical formula H₃PO₄. In water, the acid disassociates into H⁺ and H₂PO₄⁻. The second dissociation of H₂PO₄⁻⁻ results in the formation of H⁺ and HPO₄²⁻. Finally, the dissociation of HPO₄²⁻ produces H⁺ and PO₄³⁻. The following equations show the dissociation of H₃PO₄:
H₃PO₄ → H⁺ + H₂PO₄⁻
H₂PO₄⁻ → H⁺ + HPO₄²⁻
HPO₄²⁻ → H⁺ + PO₄³⁻
Using the dissociation constants of phosphoric acid, one can calculate the concentrations of all species in a 0.100 M H₃PO₄ solution. [H₃PO₄] = 0.100 M, [H₂PO₄⁻] = 0.045 M, [HPO₄²⁻] = 0.0049 M, and [PO₄³⁻] = 1.0 x 10^-7 M.
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suppose the previous forecast was 30 units, actual demand was 50 units, and ∝ = 0.15; compute the new forecast using exponential smoothing.
By using the formula of exponential smoothing, we can get the new forecast. Hence, the new forecast using exponential smoothing is 33 units.
Given:
Previous forecast = 30 units
Actual demand = 50 unitsα = 0.15Formula used:
New forecast = α(actual demand) + (1 - α)(previous forecast)
New forecast = 0.15(50) + (1 - 0.15)(30)New forecast = 7.5 + 25.5
New forecast = 33 units
Therefore, the new forecast using exponential smoothing is 33 units.
In exponential smoothing, the new forecast is computed by using the actual demand and previous forecast. In this question, the previous forecast was 30 units and actual demand was 50 units, with α = 0.15. By using the formula of exponential smoothing, we can get the new forecast. Hence, the new forecast using exponential smoothing is 33 units.
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The displacement of a wave traveling in the negative y-direction
is D(y,t)=(9.0cm)sin(45y+70t+π)D(y,t)=(9.0cm)sin(45y+70t+π), where
y is in m and t is in s.
What is the frequency of this wave?
Wh
The displacement of a wave traveling in the negative y-direction depends on the amplitude and frequency of the wave.
The displacement of a wave traveling in the negative y-direction is a combination of factors. The first factor is the amplitude, which is the maximum distance that a particle moves from its rest position as a wave passes through it. The second factor is the frequency, which is the number of waves that pass a fixed point in a given amount of time. The displacement of a wave is given by the formula y = A sin(kx - ωt + ϕ), where A is the amplitude, k is the wave number, x is the position, ω is the angular frequency, t is the time, and ϕ is the phase constant. This formula shows that the displacement depends on the amplitude and frequency of the wave.
These variables have the same fundamental meaning for waves. In any case, it is useful to word the definitions in a more unambiguous manner that applies straightforwardly to waves: Amplitude is the distance between the wave's maximum displacement and its resting position. Frequency is the number of waves that pass by a particular point every second.
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a child on a merry-go-round takes 4.4 s to go around once. what is his angular displacement during a 1.0 s time interval?
The child's angular displacement during a 1.0 s time interval is approximately 1.432 radians.
To determine the angular displacement of the child on the merry-go-round during a 1.0 s time interval, we can use the formula:
Angular Displacement (θ) = Angular Velocity (ω) × Time (t)
The angular velocity (ω) can be calculated by dividing the total angular displacement by the total time taken to complete one revolution.
In this case:
Time taken to go around once (T) = 4.4 s
Angular Velocity (ω) = 2π / T
Angular Velocity (ω) = 2π / 4.4 s ≈ 1.432 radians/s
Now, we can calculate the angular displacement during a 1.0 s time interval:
Angular Displacement (θ) = Angular Velocity (ω) × Time (t)
Angular Displacement (θ) = 1.432 radians/s × 1.0 s
Angular Displacement (θ) ≈ 1.432 radians
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The angular displacement of the child during a 1.0 s time interval is 1.44 radian. The given values are, Time taken by the child to go around once, t = 4.4 s Time interval, t₁ = 1 s
Formula used: Angular displacement (θ) = (2π/t) × t₁. Substitute the given values in the formula, Angular displacement (θ) = (2π/t) × t₁= (2π/4.4) × 1= 1.44 radian. Thus, the angular displacement of the child during a 1.0 s time interval is 1.44 radian.
The change in the angular position of an object or a point in a rotational system is known as angular displacement and it measures the amount and direction of rotation from an initial position to a final position. Angular displacement is an important concept in physics and engineering, as it helps to describe a rotational motion.
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if a dvd is spinning at 100 mph and has a radius of 14 inches, what is the linear speed of a point 3 inches from the center.
The linear speed of a point 3 inches from the center of a DVD spinning at 100 mph and with a radius of 14 inches is approximately 219.91 mph.
Linear speed is the rate at which an object moves along a circular path. It is measured in distance per unit time, such as miles per hour (mph) or meters per second (m/s).
The formula for linear speed is:
v = rω where:
v = linear speed
r = radius of the circle
rω = angular speed (measured in radians per second)
To calculate the linear speed of a point on a DVD spinning at 100 mph and with a radius of 14 inches, we need to convert the units of the given speed from mph to inches per second:
100 mph = (100 x 5280 feet) / 3600 seconds = 146.67 feet/second
146.67 feet/second = 1760 inches/second
Next, we need to find the angular speed ω of the DVD.
Angular speed is the rate at which an object rotates about an axis, and it is measured in radians per second. The formula for angular speed is:
ω = 2πf where:
ω = angular speed
f = frequency (measured in hertz)
π = 3.14159...
The frequency f of the DVD is equal to its rotational speed divided by the number of revolutions per second. One revolution is a complete turn around the circle, or 2π radians. Therefore, the frequency is:
f = (100 mph) / (2π x 14 inches x 3600 seconds/5280 feet) = 0.862 hertz
Finally, we can substitute the given values into the formula for linear speed:
v = rωv = (14 + 3) inches x 2π x 0.862 hertz = 219.91 inches/second
Therefore, the linear speed of a point 3 inches from the center of a DVD spinning at 100 mph and with a radius of 14 inches is approximately 219.91 mph.
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A 20.0-kg cannon ball is fired from a cannon with a muzzle speed of 100 m/s at an angle of 20.0° with the horizontal. Use the conservation of energy principle to find the maximum height reached by ba
A 20.0 kg cannonball is fired from a cannon with a muzzle speed of 100 m/s at an angle of 20.0°. Using conservation of energy, the maximum height reached by the cannonball is approximately 510.2 meters.
A cannon ball weighing 20.0 kg is launched from a cannon with an initial velocity of 100 m/s at an angle of 20.0° above the horizontal.
To determine the maximum height reached by the cannonball using the conservation of energy principle, we consider the conversion of kinetic energy into gravitational potential energy.
Initially, the cannonball has only kinetic energy, given by the equation KE = (1/2)mv², where m is the mass and v is the velocity.
At the highest point of its trajectory, the cannonball has no vertical velocity, meaning it has no kinetic energy but possesses gravitational potential energy, given by the equation PE = mgh, where h is the height and g is the acceleration due to gravity (approximately 9.8 m/s²).
Using the conservation of energy, we equate the initial kinetic energy to the maximum potential energy:
(1/2)mv² = mgh
Canceling the mass and rearranging the equation, we find:
v²/2g = h
Plugging in the given values, we have:
(100²)/(2*9.8) = h
Simplifying the equation, we find:
h ≈ 510.2 m
Therefore, the maximum height reached by the cannonball is approximately 510.2 meters.
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what is the best definition of relativistic thought according to perry
Relativistic thought refers to the recognition that our perceptions and beliefs are influenced by our experiences, upbringing, and cultural and social environments, according to Perry.
It suggests that reality is subjectively constructed rather than objectively discovered, and that what is "true" or "right" for one person or group may not be for another. Relativistic thinking entails a degree of tolerance for opposing viewpoints and a willingness to engage in dialogue rather than debate or dismiss opposing perspectives. Instead of seeing things in black and white, relativistic thought acknowledges the nuances and complexity of human experience and acknowledges that there may be multiple valid perspectives on any given issue.
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Water at 70 kPa and 100°C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water and the work required, in kJ/kg, for this compression. [Ans.: 664°C, 887.1 kJ/kg]
Final temperature of water is 664°C and work required for the compression process is 887.1 kJ/kg.
Given data:
Initial pressure P1 = 70 kPa
Initial temperature T1 = 100°C
Final pressure P2 = 4 MPa
Adiabatic or isentropic process, so heat transferred is zero, Q = 0
We need to determine the final temperature T2 and the work required for the compression process, W.
Adiabatic process is a process where there is no heat transfer, Q = 0. The energy balance equation for a closed system undergoing adiabatic or isentropic process can be written as:
dE = dQ - dW
Here, dE = Change in internal energy
dQ = Heat transferred (for adiabatic process, dQ = 0)
dW = Work done by the system
We can write the above equation in terms of specific quantities as: de = dq - dw
where, e = Internal energy per unit mass
q = Heat transferred per unit mass (for adiabatic process, q = 0)w = Work done per unit mass
We can use the entropy formula to determine the final temperature T2.S = constant
We can use the following equation for an adiabatic process:
S1 = S2
where S1 is the entropy of the water at P1 and T1 and S2 is the entropy of the water at P2 and T2.
S2 = S1 = constant
The entropy of the water can be calculated using the following equation:
s = Cp ln(T) - R ln(P)
where, s is the entropy per unit mass, Cp is the specific heat capacity at constant pressure, R is the gas constant, P is the pressure, and T is the temperature.
In our case, since the process is isentropic or adiabatic, the entropy change is zero.
Therefore, we can write:
S2 - S1 = 0Cp ln(T2) - R ln(P2) - Cp ln(T1) + R ln(P1) = 0Cp ln(T2/T1) - R ln(P2/P1) = 0Cp ln(T2/T1) = R ln(P1/P2)T2/T1 = (P1/P2)^(R/Cp)T2 = T1 * (P1/P2)^(R/Cp)
The specific heat capacity at constant pressure for water vapor can be taken as Cp = 1.872 kJ/kg K and the gas constant for water vapor is R = 0.4615 kJ/kg K.
The work done for an adiabatic process can be calculated using the following equation:
W = Cp * (T1 - T2)/(γ - 1)
where γ = Cp/Cv is the ratio of specific heats.
Cv for water vapor can be taken as 1.4 kJ/kg K.The specific work done per unit mass for the compression process can be calculated as:
W/m = W/m = Cp * (T1 - T2)/(γ - 1)We can substitute the given values in the above equations to obtain:
T2 = T1 * (P1/P2)^(R/Cp)T2 = 100 + 273.15 * (70 / 4000)^(0.4615/1.872) = 937.15
K = 664°CW/m = Cp * (T1 - T2)/(γ - 1)W/m = 1.872 * (100 + 273.15 - 937.15)/(1.4 - 1) = -887.1 kJ/kg
Work required for the compression process is 887.1 kJ/kg.
Final temperature of water is 664°C and work required for the compression process is 887.1 kJ/kg.
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a lens has a refractive power of -1.50. what is its focal length?
It has been determined that the focal length of the lens is -0.6667 m.
Given: The refractive power of a lens is -1.50We are supposed to find the focal length of the given lens
Solution:The formula to find the focal length of a lens is given by:1/f = (n-1) (1/R1 - 1/R2)
Given: Refractive power (P) = -1.50
As we know that, P = 1/f (Where f is the focal length)
Hence, -1.50 = 1/fOr, f = -1/1.5= -0.6667 m
Therefore, the focal length of the given lens is -0.6667 m.
From the above calculations, it has been determined that the focal length of the lens is -0.6667 m.
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21.42 using cyclopentanone as your starting material and using any other reagents of your choice, propose an efficient synthesis for each of the following compounds
Cyclopentanone, C5H8O is a cyclic ketone and can be converted to various organic compounds with the help of different reagents. Thus, cyclopentanone can be used as a starting material to synthesize different organic compounds using various reagents and catalysts.
Here, efficient syntheses for three organic compounds using cyclopentanone as a starting material are given below:
1) 2-Methylcyclopentanone: It can be prepared by the reaction of cyclopentanone with isopropyl, magnesium bromide, followed by hydrolysis of the resulting product. This reaction is shown below:
2) Cyclopentylmethanol: It can be prepared by the reduction of cyclopentanone with sodium borohydride (NaBH4) in methanol. This reaction is shown below:
3) 2-Cyclopenten-1-one: It can be prepared by the dehydration of cyclopentanol, which can be prepared by the reduction of cyclopentanone with lithium aluminum hydride (LiAlH4). The dehydration of cyclopentanol can be carried out by the elimination of water molecule using an acid catalyst like H2SO4. The overall reaction is shown below.
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Question 1 Calculate the amount of radiation emitted by a blackbody with a temperature of 353 K. Round to the nearest whole number (e.g., no decimals) and input a number only, the next question asks a
The amount of radiation emitted by a blackbody with a temperature of 353 K is 961 {W/m}².
The formula for calculating the amount of radiation emitted by a blackbody is given by the Stefan-Boltzmann law: j^* = \sigma T^4 Where j* is the radiation energy density (in watts per square meter), σ is the Stefan-Boltzmann constant (σ = 5.67 x 10^-8 W/m^2K^4), and T is the absolute temperature in Kelvin (K).Using the given temperature of T = 353 K and the formula above, we can calculate the amount of radiation emitted by the blackbody: j^* = \sigma T^4 j^* = (5.67 \times 10^{-8}) (353)^4 j^* = 961.2 {W/m}².
Therefore, the amount of radiation emitted by the blackbody with a temperature of 353 K is approximately 961 watts per square meter (W/m²).Rounding this to the nearest whole number as specified in the question gives us the final answer of: 961 (no decimals).
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A capacitor is discharged through a 20.0 Ω resistor. The discharge current decreases to 22.0% of its initial value in 1.50 ms.
What is the time constant (in ms) of the RC circuit?
a) 0.33 ms
b) 0.67 ms
c) 1.50 ms
d) 3.75 ms
The time constant (in ms) of the RC circuit is 3.75 ms. Hence, the correct option is (d) 3.75 ms.
The rate of decay of the current in a charging capacitor is proportional to the current in the circuit at that time. Therefore, it takes longer for a larger current to decay than for a smaller current to decay in a charging capacitor.A capacitor is discharged through a 20.0 Ω resistor.
The discharge current decreases to 22.0% of its initial value in 1.50 ms. We can obtain the time constant of the RC circuit using the following formula:$$I=I_{o} e^{-t / \tau}$$Where, I = instantaneous current Io = initial current t = time constant R = resistance of the circuit C = capacitance of the circuit
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The time constant of the RC circuit is approximately 0.674 m s.
To determine the time constant (τ) of an RC circuit, we can use the formula:
τ = RC
Given that the discharge current decreases to 22.0% of its initial value in 1.50 m s, we can calculate the time constant as follows:
The percentage of the initial current remaining after time t is given by the equation:
I(t) =[tex]I_oe^{(-t/\tau)[/tex]
Where:
I(t) = current at time t
I₀ = initial current
e = Euler's number (approximately 2.71828)
t = time
τ = time constant
We are given that the discharge current decreases to 22.0% of its initial value. Therefore, we can set up the following equation:
0.22 =[tex]e^{(-1.50/\tau)[/tex]
To solve for τ, we can take the natural logarithm (ln) of both sides:
ln(0.22) = [tex]\frac{-1.50}{\tau}[/tex]
Rearranging the equation to solve for τ:
τ = [tex]\frac{-1.50 }{ ln(0.22)}[/tex]
Calculating this expression:
τ ≈ 0.674 m s
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A charge -5.5 nC is placed at (-3.1.-3) m and another charge 9.3 nC is placed at (-2,3,-2) m. What is the electric field at (1,0,0)m?
The electric field at (1,0,0) m due to the given charges is -1.2 x 10^5 N/C, directed towards the left.
Let's first calculate the electric field at point P due to the first charge:q1 = -5.5 nC, r1 = (-3.1, -3, 0) m and r = (1, 0, 0) m
The distance between charge 1 and point P is:r = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)²)r = √((1 - (-3.1))² + (0 - (-3))² + (0 - 0)²)r = √(4.1² + 3² + 0²)r = 5.068 m
Therefore, the electric field at point P due to charge 1 is:
E1 = kq1 / r1²E1 = (9 x 10^9 Nm²/C²) x (-5.5 x 10^-9 C) / (5.068 m)²E1 = -4.3 x 10^5 N/C (towards left, as the charge is negative)
Now, let's calculate the electric field at point P due to the second charge:
q2 = 9.3 nC, r2 = (-2, 3, -2) m and r = (1, 0, 0) m
The distance between charge 2 and point P is:
r = √((x2 - x1)² + (y2 - y1)² + (z2 - z1)²)
r = √((1 - (-2))² + (0 - 3)² + (0 - (-2))²)
r = √(3² + 3² + 2²)r = √22 m
Therefore, the electric field at point P due to charge 2 is:
E2 = kq2 / r2²
E2 = (9 x 10^9 Nm²/C²) x (9.3 x 10^-9 C) / (√22 m)²
E2 = 3.1 x 10^5 N/C (towards right, as the charge is positive)
Now, the total electric field at point P due to both charges is:
E = E1 + E2
E = -4.3 x 10^5 N/C + 3.1 x 10^5 N/C
E = -1.2 x 10^5 N/C
Therefore, the electric field at (1,0,0) m due to the given charges is -1.2 x 10^5 N/C, directed towards the left.
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The electric field at point P (1, 0, 0)m is (-2.42 × 10⁶) î + 6.91 × 10⁶ ĵ N/C.
The given charges are -5.5 nC and 9.3 nC. The position vectors of these charges are (-3.1, -3, 0)m and (-2, 3, -2)m. We need to find the electric field at (1, 0, 0)m.
Let's consider charge q1 (-5.5 nC) and charge q2 (9.3 nC) respectively with position vectors r1 and r2. Electric field due to q1 at point P (1,0,0)m is given by:r1 = (-3.1, -3, 0)mq1 = -5.5 nC
Position vector r from q1 to P = rP - r1 = (1, 0, 0)m - (-3.1, -3, 0)m = (4.1, 3, 0)m
Using the formula of electric field, the electric field due to q1 at point P will be given by:
E1 = kq1 / r²
where k is the Coulomb constantk = 9 × 10⁹ N m² C⁻²
Electric field due to q1 at point P isE1 = 9 × 10⁹ × (-5.5) / (4.1² + 3²) = -2.42 × 10⁶ N/C
Now, let's consider charge q2. The position vector of q2 is given by:r2 = (-2, 3, -2)mq2 = 9.3 nC
Position vector r from q2 to P = rP - r2 = (1, 0, 0)m - (-2, 3, -2)m = (3, -3, 2)m
Electric field due to q2 at point P will be given by:
E2 = kq2 / r²
Electric field due to q2 at point P is
E2 = 9 × 10⁹ × 9.3 / (3² + (-3)² + 2²) = 6.91 × 10⁶ N/C
Now, we can get the total electric field due to the given charges by adding the electric fields due to q1 and q2 vectorially.
The vector addition of electric fields E1 and E2 is given by the formula:
E = E1 + E2
Let's consider charge q1 (-5.5 nC) and charge q2 (9.3 nC) respectively with position vectors r1 and r2. Electric field due to q1 at point P (1,0,0)m is given by:r1 = (-3.1, -3, 0)mq1 = -5.5 nC
Position vector r from q1 to P = rP - r1 = (1, 0, 0)m - (-3.1, -3, 0)m = (4.1, 3, 0)m
Using the formula of electric field, the electric field due to q1 at point P will be given by:E1 = kq1 / r²
where k is the Coulomb constant
k = 9 × 10⁹ N m² C⁻²
The magnitude of the electric field due to q1 at point P is given by|E1| = 9 × 10⁹ × |q1| / r²= 9 × 10⁹ × 5.5 / (4.1² + 3²) N/C= 2.42 × 10⁶ N/C
The direction of the electric field due to q1 at point P is towards the charge q1.
Now, let's consider charge q2. The position vector of q2 is given by:r2 = (-2, 3, -2)mq2 = 9.3 nC
Position vector r from q2 to P = rP - r2 = (1, 0, 0)m - (-2, 3, -2)m = (3, -3, 2)m
The magnitude of the electric field due to q2 at point P will be given by:
E2 = kq2 / r²= 9 × 10⁹ × 9.3 / (3² + (-3)² + 2²) N/C= 6.91 × 10⁶ N/C
The direction of the electric field due to q2 at point P is away from the charge q2.
Now, we can get the total electric field due to the given charges by adding the electric fields due to q1 and q2 vectorially. The vector addition of electric fields E1 and E2 is given by the formula:E = E1 + E2E = (-2.42 × 10⁶) î + 6.91 × 10⁶ ĵ N/C
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A 0.200-kg object is attached to a spring that has a force constant of 95.0 N/m. The object is pulled 7.00 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table. Calculate the maximum speed Umas of the object. Upis m/y Find the location x of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up. m
The maximum speed of the object is Umas = 1.516 m/s. The location of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up is x = 6.97 cm..
To find the maximum speed of the object, we can use the concept of mechanical energy conservation. At the maximum speed, all the potential energy stored in the spring is converted into kinetic energy.
The potential energy stored in the spring is given by:
Potential energy (PE) = (1/2)kx²
Where:
k = force constant of the spring = 95.0 N/m
x = displacement of the object from equilibrium = 7.00 cm = 0.0700 m (converted to meters)
Substituting the values into the equation:
PE = (1/2)(95.0 N/m)(0.0700 m)²
PE ≈ 0.230 Joules
At the maximum speed, all the potential energy is converted into kinetic energy:
Kinetic energy (KE) = 0.230 Joules
The kinetic energy is given by:
KE = (1/2)mv²
Where:
m = mass of the object = 0.200 kg
v = maximum speed of the object (Umas)
Substituting the values into the equation:
0.230 Joules = (1/2)(0.200 kg)v²
v² = (0.230 Joules) * (2/0.200 kg)
v² = 2.30 Joules/kg
v ≈ 1.516 m/s
Therefore, the maximum speed of the object is Umas ≈ 1.516 m/s.
To find the location of the object relative to equilibrium when it has one-third of the maximum speed, we can use the concept of energy conservation again. At this point, the kinetic energy is one-third of the maximum kinetic energy.
KE = (1/2)mv²
(1/3)KE = (1/6)mv²
Substituting the values into the equation:
(1/3)(0.230 Joules) = (1/6)(0.200 kg)v²
0.077 Joules = (0.0333 kg)v²
v² = 2.311 Joules/kg
v ≈ 1.519 m/s
Now, we need to find the displacement x of the object from equilibrium at this velocity. We can use the formula for the potential energy stored in the spring:
PE = (1/2)kx²
Rearranging the equation:
x² = (2PE) / k
x² = (2 * 0.230 Joules) / 95.0 N/m
x² ≈ 0.004842 m²
x ≈ ±0.0697 m
Since the object is moving to the right, the displacement x will be positive:
x ≈ 0.0697 m
Converting this to centimeters:
x ≈ 6.97 cm
Therefore, the location of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up is x ≈ 6.97 cm.
The maximum speed of the object is Umas ≈ 1.516 m/s. The location of the object relative to equilibrium when it has one-third of the maximum speed, is moving to the right, and is speeding up is x ≈ 6.97 cm.
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