A 2.00−kg
, frictionless block is attached to an ideal spring with force constant 300 N/m
. At t=0
the spring is neither stretched nor…
A 2.00−kg
, frictionless block is attached to an ideal spring with force constant 300 N/m
. At t=0
the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s
. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

Answers

Answer 1

Using the equation for simple harmonic motion, we find that (a) The amplitude A is zero. (b) The phase angle φ is zero. (c) The equation for the position x(t) as a function of time is x(t) = 0.

To find the amplitude and phase angle of the block attached to the spring, we can use the equation for simple harmonic motion:

x(t) = A * cos(ωt + φ),

where:

x(t) is the position of the block as a function of time,A is the amplitude,ω is the angular frequency, andφ is the phase angle.We can start by finding the angular frequency ω using the formula:

ω = √(k/m),

where k is the force constant of the spring and m is the mass of the block. Plugging in the given values:

k = 300 N/m,

m = 2.00 kg.

ω = √(300 N/m / 2.00 kg) = √(150 N/kg) ≈ 12.25 rad/s.

Now, let's find the amplitude A. The amplitude represents the maximum displacement from the equilibrium position. In this case, the block starts at rest (neither stretched nor compressed), moving in the negative direction at 12.0 m/s.

Since it's already moving in the negative direction, we know that the amplitude will also be negative.

v(t) = -A * ω * sin(ωt + φ).

At t = 0, v(0) = -12.0 m/s, so we can write:

-12.0 m/s = -A * ω * sin(0 + φ).

Since sin(0) = 0, we have:

-12.0 m/s = 0.

Therefore, the amplitude A is zero in this case.

Next, let's find the phase angle φ. We can use the formula:

v(t) = -A * ω * sin(ωt + φ).

At t = 0, v(0) = -12.0 m/s, so we can write:

-12.0 m/s = -A * ω * sin(0 + φ).

Since sin(0) = 0, the equation simplifies to:

-12.0 m/s = 0.

Therefore, the phase angle φ is also zero in this case.

Finally, we can write the equation for the position x(t) as a function of time:

x(t) = A * cos(ωt + φ).

Since the amplitude A and phase angle φ are both zero, the equation simplifies to:

x(t) = 0.

So, the position of the block as a function of time is always zero. This means the block remains at its equilibrium position throughout the motion and does not oscillate.

In summary:

(a) The amplitude A is zero.

(b) The phase angle φ is zero.

(c) The equation for the position x(t) as a function of time is x(t) = 0.

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Related Questions

. Two blocks are connected by a light string that passes over a frictionless pulley, as shown in figure P12. The block of mass m1 lies on a horizontal surface and is connected to a spring of force constant k. The system is released from rest when the spring is unstretched. If the hanging block of mass m2 falls a distance h before coming to rest, calculate the coefficient of kinetic friction between the block of mass m1 and the surface

Answers

The coefficient of kinetic friction between the block of mass m1 and the surface is given by ([tex]mg_2[/tex] - kh) / [tex]mg_1[/tex].

Let's assume that m1 is the mass on the horizontal surface and [tex]m_2[/tex] is the hanging mass.

The forces acting on m1 are the force of gravity ([tex]mg_1[/tex]) and the force of friction ([tex]f_ {friction[/tex]).

The force of gravity acting on [tex]m_2[/tex] is [tex]mg_2[/tex].

[tex]mg_2[/tex] = T -- (Equation 1)

The tension in the string (T) is pulling m1 towards the pulley, causing the block to move. Therefore, we can write:

T = [tex]f_{friction}[/tex] + [tex]mg_1[/tex] -- (Equation 2)

Here, x represents the displacement of m1 from the equilibrium position. So, we can write:

kx =[tex]f_{friction[/tex]-- (Equation 3)

Now, let's substitute Equation 3 with Equation 2:

T = kx + [tex]mg_1[/tex]

From Equation 1, we know that T = [tex]mg_2[/tex].

Substituting this into the above equation:

[tex]mg_2[/tex] = kx + [tex]mg_1[/tex]

Since the hanging block falls a distance h, we can write h = x. Substituting this into the equation:

[tex]mg_2[/tex] = kh + [tex]mg_2[/tex]

Now, let's solve for the coefficient of kinetic friction (μ_k).  So, we can write:

[tex]f_{friction} = \mu _{kmg1[/tex]

Substituting this into the equation:

[tex]mg_2[/tex] = kh +[tex]\mu \ _{kmg1}[/tex]

Simplifying the equation, we get:

[tex]\mu \ _{k}[/tex]= ([tex]mg_2[/tex] - kh) / [tex]mg_1[/tex]

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if a rock band turned up its amplifiers to produce twice as much energy as it produced before, the listener would experience a sound twice as loud
t/F

Answers

True, doubling the energy produced by the amplifiers of a rock band will result in the listener experiencing a sound that is approximately twice as loud.

How does sound perception vary?

Sound intensity is directly related to the perceived loudness by the listener. When a rock band turns up its amplifiers to produce twice as much energy, it results in an increase in the sound intensity.

According to the psychoacoustic perception of sound, the relationship between sound intensity and perceived loudness is not linear but logarithmic.

Doubling the energy output of the amplifiers roughly corresponds to an increase of approximately 3 decibels (dB), which is perceived as a noticeable increase in loudness.

This increase is due to the logarithmic nature of the decibel scale, where a 3 dB increase represents a doubling of sound intensity.

However, it is important to note that individual perception of loudness can vary, and factors such as distance from the sound source, the environment, and personal hearing sensitivity can also influence the subjective experience of sound.

In conclusion, doubling the energy produced by the amplifiers of a rock band generally results in a sound that is perceived as approximately twice as loud by the listener, taking into account the logarithmic relationship between sound intensity and perceived loudness.

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A swimming pool is nearly empty, holding only 5300 gallons of water. A system is set up so that the water in the pool starts to increase by 16% per hour. After how many hours is the pool filling at a rate of 2657 gallons per hour?The pool is filling at a rate of 2657 gallons per hour after nothing hours. If necessary, round to two decimal places.

Answers

The pool is filling at a rate of 2657 gallons per hour after 0.3125 hours (18.75 minutes).

Given, A swimming pool is nearly empty, holding only 5300 gallons of water. The water in the pool starts to increase by 16% per hour. We need to find how many hours will it take to fill the pool at a rate of 2657 gallons per hour.

We know that the pool starts filling up by 16% per hour.Let the rate of increase be x gallons per hour.

So, the amount of water in pool after 1 hour = 5300 + x

After 2 hours = 5300 + 2x

So on, After n hours, amount of water in the pool = 5300 + nx

From the above data, we can derive a formula for the amount of water in the pool after n hours.5300(1 + 0.16n)We know that the pool is filling up at a rate of 2657 gallons per hour.

So, 2657 = 5300(0.16n + 1)0.16n + 1 = 2657/53000.16n + 1 = 0.05n = 0.05 / 0.160.05 / 0.16 = 0.3125 hours

Therefore, it will take 0.3125 hours or 18.75 minutes to fill the pool at a rate of 2657 gallons per hour. Thus, the pool is filling at a rate of 2657 gallons per hour after 0.3125 hours (18.75 minutes).

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Assume that at time tmtmt_m, the current in the inductor is at a maximum; at that time, the current flows from point b to point a. At time tmtmt_m, which of the following statements is true?
The voltage across the inductor must be zero and increasing.
The voltage across the inductor must be zero and decreasing.
The voltage across the inductor must be positive and momentarily constant.
The voltage across the inductor must be negative and momentarily constant.

Answers

When the current in the inductor is at a maximum at time tmtm, the voltage across the inductor must be zero and the rate of change of the current is maximum. Therefore, the correct option is (b) The voltage across the inductor must be zero and decreasing.

The inductor is a passive electrical component that stores energy in a magnetic field produced by the current flowing through it. When current flows through an inductor, it generates a magnetic field proportional to the current's rate of change. The magnetic field creates an opposing voltage that opposes the change in current flow in the inductor.When the current in the inductor is at a maximum at time tmtm, the rate of change of current is zero. Hence the opposing voltage across the inductor is zero. As the current continues to flow, the magnetic field is dissipated, and the inductor's voltage returns to its original value. The current continues to flow through the circuit. However, the rate of change of the current is now negative, causing the magnetic field to begin to collapse. This change in magnetic field polarity generates an opposing voltage that opposes the current's flow in the circuit. As a result, the voltage across the inductor is zero and decreasing at time tmtm. Hence, option (b) is correct.

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a 13.0 µf capacitor is charged by a 10.0 v battery through a resistance r. the capacitor reaches a potential difference of 4.00 v at a time 3.00 s after charging begins. find r. kω

Answers

The resistance, r, in kiloohms (kΩ), is approximately 3.33 kΩ.
We can use the formula for the charging of a capacitor to find the resistance, given the capacitance, voltage, and time. The formula is:

V = V0(1 - e^(-t/RC))

Where:
V is the final voltage across the capacitor (4.00 V),
V0 is the initial voltage across the capacitor (10.0 V),
t is the time (3.00 s),
R is the resistance (in ohms),
and C is the capacitance (13.0 µF).

Plugging in the known values, we get:

4.00 V = 10.0 V(1 - e^(-3.00 s/(R * 13.0 × 10^(-6) F)))

To solve for R, we rearrange the equation:

e^(-3.00 s/(R * 13.0 × 10^(-6) F)) = 1 - (4.00 V / 10.0 V)

e^(-3.00 s/(R * 13.0 × 10^(-6) F)) = 0.6

Taking the natural logarithm (ln) of both sides:

-3.00 s/(R * 13.0 × 10^(-6) F) = ln(0.6)

Solving for R:

R = -3.00 s / (13.0 × 10^(-6) F * ln(0.6))

R ≈ 3.33 kΩ

The resistance, r, required to charge the 13.0 µF capacitor to a potential difference of 4.00 V in 3.00 seconds is approximately 3.33 kΩ.

In conclusion, to charge a 13.0 µF capacitor with a 10.0 V battery to a potential difference of 4.00 V within a time of 3.00 seconds, a resistance of approximately 3.33 kΩ is needed. This result was obtained by utilizing the formula for the charging of a capacitor, which relates the voltage across the capacitor to the initial voltage, time, resistance, and capacitance.

By rearranging the formula and substituting the known values, we obtained an equation involving the exponential function. Taking the natural logarithm of both sides allowed us to isolate the variable R, representing the resistance. Solving the equation yielded a value of approximately 3.33 kΩ for the resistance.

It is important to note that the calculated resistance value is an approximation due to rounding. However, it provides a close estimate that fulfills the given conditions. The resistance value serves as a guide for selecting an appropriate resistor to achieve the desired charging behavior in the circuit.

Therefore, to charge the capacitor under the specified conditions, a resistance of approximately 3.33 kΩ should be employed.

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light visible to humans consists of electromagnetic waves with wavelengths (in air) in the range 400-700 nm (4.0 10-7 m to 7.0 10-7 m). the speed of light in air is 3.0 108 m/s. what are the frequencies of electromagnetic waves that are visible? (enter your answers from smallest to largest.)

Answers

The frequencies of visible electromagnetic waves range from approximately 4.3 x 10¹⁴ Hz to 7.5 x 10¹⁴ Hz.

To calculate the frequencies of electromagnetic waves that are visible, we can use the formula:

frequency = speed of light / wavelength

Given:

Speed of light in air = 3.0 × 10⁸ m/s

Wavelength range in air = 400 nm to 700 nm (4.0 × 10⁻⁷ m to 7.0 × 10⁻⁷m)

Step 1: Convert the wavelength range from nanometers (nm) to meters (m):

Minimum wavelength = 400 nm = 400 × 10⁻⁹ m = 4.0 × 10⁻⁷ m

Maximum wavelength = 700 nm = 700 × 10⁻⁹ m = 7.0 × 10⁻⁷ m

Step 2: Calculate the frequencies using the formula:

Minimum frequency = speed of light / minimum wavelength

Maximum frequency = speed of light / maximum wavelength

Minimum frequency = (3.0 × 10⁸ m/s) / (4.0 × 10⁻⁷ m)

Minimum frequency ≈  7.5 x 10¹⁴Hz

Maximum frequency = (3.0 × 10₈ m/s) / (7.0 × 10⁻⁷ m)

Maximum frequency ≈  4.3 x 10¹⁴ Hz

So, the frequencies of electromagnetic waves that are visible range from approximately 4.3 x 10¹⁴ Hz to  7.5 x 10¹⁴ Hz.

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Why is there no work done in moving a charge from one point to another on an equipotential surface

Answers

Work is not done in moving a charge on an equipotential surface because the potential remains constant throughout the surface.

Why does the potential remain constant on an equipotential surface?

When a charge is moved from one point to another on an equipotential surface, no work is done because the potential at every point on the surface is the same. An equipotential surface is a region in space where the electric potential has a constant value.

This means that the work done in moving a charge along an equipotential surface is zero because there is no change in potential energy. The electric field lines on an equipotential surface are perpendicular to the surface, and any movement of a charge occurs parallel to the field lines.

As a result, the force exerted on the charge is always perpendicular to the direction of motion, resulting in no work done.

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A spherical balloon is inflating with helium at of 180π ft^3/min. How fast is the balloon's radius increasing at the instant the radius is 3 ft? How fast is the surface area increasing ?
The balloon's radius is increasing at rate of ___ ft/min at instant the radius is 3 ft

Answers

The balloon's radius is increasing at a rate of 60 ft/min at the instant the radius is 3 ft. The surface area of the balloon is increasing at a rate of 360π ft²/min.

To find how fast the balloon's radius is increasing, we can use the formula for the volume of a sphere:

V = (4/3)πr³

where V is the volume and r is the radius of the sphere. We are given that the volume is increasing at a rate of 180π ft³/min. We can differentiate the volume equation with respect to time to find the rate of change of the volume:

dV/dt = (4/3)π(3r²(dr/dt))

Since we know dV/dt is 180π ft³/min and r is 3 ft (at the instant the radius is 3 ft), we can solve for (dr/dt):

180π = (4/3)π(3²(dr/dt))

180 = 12(dr/dt)

dr/dt = 180/12 = 15 ft/min

Therefore, the balloon's radius is increasing at a rate of 15 ft/min when the radius is 3 ft.

To find how fast the surface area of the balloon is increasing, we can differentiate the formula for the surface area of a sphere:

A = 4πr²

where A is the surface area and r is the radius.

Differentiating with respect to time gives:

dA/dt = 8πr(dr/dt)

We already know dr/dt is 15 ft/min and r is 3 ft, so we can substitute these values into the equation:

dA/dt = 8π(3)(15)

dA/dt = 360π

Therefore, the surface area of the balloon is increasing at a rate of 360π ft²/min.

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an object at rest begins to rotate with a constant angular acceleration. if this object has angular velocity ω at time t, what was its angular velocity at the time ½t?

Answers

The angular velocity at time ½t is half of the angular velocity at time t.

When an object rotates with a constant angular acceleration, its angular velocity increases linearly with time. We can express this relationship using the equation:

ω = ω0 + αt,

where ω is the final angular velocity at time t, ω0 is the initial angular velocity at time 0, α is the angular acceleration, and t is the time.

If we substitute t/2 for t in the equation, we get:

ω(1/2) = ω0 + α(t/2),

Simplifying further:

ω(1/2) = ω0 + (α/2)t,

Since α is the constant angular acceleration, we can rewrite it as α = (ω - ω0)/t. Substituting this into the equation above, we have:

ω(1/2) = ω0 + ((ω - ω0)/2),

Simplifying gives:

ω(1/2) = (ω + ω0)/2.

Therefore, at time ½t , the angular velocity is half of that at time t.

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the process of heating a pot of water from room temperature to boilling temperature to boilling temperature is an endothermic process
T/F

Answers

The given statement "The process of heating a pot of water from room temperature to boiling temperature is an endothermic process" is true because it accurately reflects the thermodynamics involved in heating water.

Is the process of heating water endothermic?

When heating a pot of water from room temperature to boiling temperature, the process is indeed endothermic. In an endothermic process, energy is absorbed from the surroundings to increase the internal energy of the system.

In this case, heat energy is transferred to the water molecules, causing them to gain kinetic energy and eventually reach the boiling point. As the water absorbs heat, it undergoes a phase change from a liquid to a gas. This requires energy input, resulting in an endothermic process.

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if the system mass is [var:m_sys] grams and the mass of the disks hanging on the string is [var:m_hang] grams, predict the acceleration.

Answers

If the system mass is 600grams and the mass of the disks hanging on the string is100grams, acceleration of the system is approximately -8.17 m/s².

To predict the acceleration of the system, we need to consider the forces acting on it. In this case, there are two forces involved: the force due to gravity (weight) and the tension in the string.

The force due to gravity can be calculated using the equation:

Weight = mass * gravity

Where mass is the total mass of the system (600 grams = 0.6 kg) and gravity is the acceleration due to gravity (approximately 9.8 m/s²).

Weight = 0.6 kg * 9.8 m/s² = 5.88 N

The tension in the string is equal to the weight of the disks hanging on it, which is given as 100 grams (0.1 kg) in this case.

Tension = 0.1 kg * 9.8 m/s² = 0.98 N

Now, let's use Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration:

Net force = mass * acceleration

The net force is the difference between the tension in the string and the weight of the system:

Net force = Tension - Weight

Net force = 0.98 N - 5.88 N = -4.9 N (Note: The negative sign indicates that the net force is in the opposite direction to the weight)

Substituting the values into Newton's second law, we have:

-4.9 N = 0.6 kg * acceleration

Solving for acceleration:

acceleration = (-4.9 N) / (0.6 kg) ≈ -8.17 m/s²

The negative sign indicates that the system is experiencing a deceleration or slowing down.

Therefore, the predicted acceleration of the system is approximately -8.17 m/s².

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what is the maximum amount by which the wavelength of an incident photon could change when it undergoes compton scattering from an atom or molecule with a 24.0 u mass? (1 u = 1.66 × 10-27 kg).

Answers

An atom or molecule with a mass of 24.0 u can modify the wavelength of an incident photon during Compton scattering by a maximum of 2.43 × 10⁻¹² meters.

To calculate the maximum change in wavelength during Compton scattering, we can use the formula derived from the Compton scattering equation:

[tex]\begin{equation}\Delta \lambda = \lambda' - \lambda = \frac{h}{m_0 c}(1 - \cos \theta)[/tex]

Where:

Δλ is the change in wavelength,

λ' is the final wavelength after scattering,

λ is the initial wavelength,

h is the Planck's constant (6.626 × 10⁻³⁴ J·s),

m₀ is the rest mass of the electron (9.11 × 10⁻³¹ kg),

c is the speed of light (3.00 × 10⁸ m/s),

θ is the scattering angle.

Given that the rest mass of an atom or molecule is 24.0 u (1 u = 1.66 × 10⁻²⁷ kg), we need to convert it to kilograms:

m₀ = 24.0 u * 1.66 × 10⁻²⁷ kg/u

   = 3.984 × 10⁻²⁶ kg

To find the maximum change in wavelength, we need to consider the scattering angle that yields the maximum change. This occurs when the photon is scattered at a 180-degree angle (backscattering).

θ = 180 degrees = π radians

Substituting the values into the formula:

[tex]\begin{equation}\Delta \lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(3.984 \times 10^{-26} \text{ kg})(3.00 \times 10^{8} \text{ m/s})}(1 - \cos \pi)[/tex]

   ≈ 2.43 × 10⁻¹² m

Therefore, the maximum amount by which the wavelength of an incident photon could change during Compton scattering from an atom or molecule with a 24.0 u mass is approximately 2.43 × 10⁻¹² meters.

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ball 1, with a mass of 100 g and traveling at 12.0 m/s , collides head on with ball 2, which has a mass of 330 g and is initially at rest.
Q1: What is the final velocity of the ball 1 AND ball 2 if the collision is perfectly elastic? Q2: What is the final velocity of the ball 1 AND 2 if the collision is perfectly inelastic?

Answers

Q1.  -6.18 m/s is the final velocity of the ball 1 AND ball 2 if the collision is perfectly elastic. Q2. 2.72 m/s  is the final velocity of the ball 1 AND 2 if the collision is perfectly inelastic.

Q1: In a perfectly elastic collision, both momentum and kinetic energy are conserved. Applying the conservation of momentum, we can calculate the final velocities of the balls. Let v1 and v2 represent the final velocities of ball 1 and ball 2, respectively. Using the equation for conservation of momentum, [tex]m1v1 + m2v2 = m1u1 + m2u2[/tex],

where m1 and m2 are the masses of ball 1 and ball 2, and u1 and u2 are their initial velocities. Given the values of the masses and initial velocities, we can solve for v1 and v2.

100×12=330×V

V=-6.18 m/s

Q2: In a perfectly inelastic collision, the two balls stick together after the collision and move as a single unit. The principle of conservation of momentum still applies, but kinetic energy is not conserved. Again, using the equation for conservation of momentum, [tex]m1v1 + m2v2 = (m1 + m2)v[/tex], where v is the final velocity of the combined system. Given the masses and initial velocities, we can solve for v to determine the final velocity of the combined system (which will be the same for both balls).

100×12=380×V

V= 2.72 m/s

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A 22 cm × 22 cm square loop has a resistance of 0.20 Ω . A magnetic field perpendicular to the loop is B=4t−2t2, where B is in tesla and t is in seconds.
PART A: What is the current in the loop at t=0.0s?
PART B: What is the current in the loop at t=1.0s?
PART C: What is the current in the loop at t=2.0s?

Answers

The current in the loop at t=2.0s is (4 V)/(0.20 Ω) = 20 Ampere.

PART A: At t=0.0s, the magnetic field B is given as B=4(0)-2(0)^2 = 0 Tesla. Since there is no magnetic field passing through the loop at t=0.0s, there is no change in the magnetic flux. Therefore, according to Faraday's law of electromagnetic induction, no electromotive force is induced, and hence, there is no current flowing through the loop. Therefore, the current in the loop at t=0.0s is 0 Ampere.

PART B: At t=1.0s, the magnetic field B is given as B=4(1)-2(1)^2 = 2 Tesla. With a magnetic field of 2 Tesla passing through the loop, a change in magnetic flux occurs. According to Faraday's law, an electromotive force is induced in the loop. The induced electromotive force is given by the equation EMF = -dΦ/dt, where dΦ/dt represents the rate of change of magnetic flux.

As the loop has a resistance of 0.20 Ω, the induced electromotive force causes a current to flow through the loop. Using Ohm's law, we can calculate the current by dividing the induced electromotive force by the resistance. Therefore, the current in the loop at t=1.0s is (2 V)/(0.20 Ω) = 10 Ampere.

PART C: At t=2.0s, the magnetic field B is given as B=4(2)-2(2)^2 = 4 Tesla. Similar to the previous case, a change in magnetic flux occurs with a magnetic field of 4 Tesla passing through the loop. The induced electromotive force is given by EMF = -dΦ/dt, and the current can be calculated using Ohm's law. Therefore, the current in the loop at t=2.0s is (4 V)/(0.20 Ω) = 20 Ampere.

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a resistor and an inductor are connected in series to an ideal battery of constant terminal voltage. at the moment contact is made with the battery, the voltage across the resistor is

Answers

At the moment contact is made with the battery, the voltage across the resistor is equal to the terminal voltage of the battery.

In a series circuit, the same current flows through each component. When a battery is connected to a resistor and an inductor in series, the current starts to flow through the circuit. However, in the case of an inductor, it opposes changes in current by inducing an opposing voltage.

At the moment contact is made with the battery, the inductor opposes the sudden change in current by generating a back EMF (electromotive force). This back EMF cancels out the initial voltage drop across the resistor.

Therefore, the voltage across the resistor is equal to the terminal voltage of the battery at the moment of contact. This is because the inductor needs some time to build up its magnetic field and fully oppose the change in current.

At the moment contact is made with the battery, the voltage across the resistor is equal to the terminal voltage of the battery. The inductor needs time to generate its opposing voltage, so initially, the voltage drop across the resistor is equal to the battery's terminal voltage.

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True/False. bending moment is maximum where shear force passes through zero

Answers

True. The bending moment is maximum where the shear force passes through zero.

The bending moment and shear force are related to the internal forces within a structure. The shear force represents the internal forces that cause a structure to slide or shear along a specific section, while the bending moment represents the internal forces that cause the structure to bend or deform.

When the shear force passes through zero at a particular section, it means that the internal forces are changing direction, from compressive to tensile or vice versa. At this point, the bending moment is at its maximum.

To understand why this is true, consider a beam subjected to an applied load. As the load moves along the beam, the shear force changes sign when it passes through zero, indicating a change in the direction of internal forces. At this point, the bending moment reaches its maximum value because the forces are transitioning from one type of internal force to another.

Therefore, it can be concluded that the bending moment is indeed maximum where the shear force passes through zero.

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hfc-134a has a global warming potential (gwp) of:

Answers

Answer:

HFC-134a, also known as R-134a, has a global warming potential (GWP) of 1.

Explanation:

The GWP is a measure of how much a given greenhouse gas contributes to global warming over a specific time period compared to carbon dioxide (CO2), which has a GWP of 1. HFC-134a has a GWP of 1 over a time horizon of 100 years. This means that, in terms of global warming potential, it has a relatively low impact compared to some other greenhouse gases, such as methane (GWP of 28-36 over 100 years) or hydrofluorocarbons with higher GWPs.  

However, it's important to note that HFC-134a is still considered a greenhouse gas and contributes to global warming. Its use as a refrigerant and in air conditioning systems has been a concern due to its potential environmental impact. Efforts are being made to phase out and replace HFC-134a with more environmentally friendly alternatives to mitigate its contribution to climate change.

HFC-134a has a global warming potential (GWP) of 1,430.The global warming potential (GWP) is a measure of how much a particular greenhouse gas contributes to global warming over a specific timeframe compared to carbon dioxide (CO2). CO2 is assigned a GWP of 1, and other gases are given GWPs relative to CO2.

HFC-134a, also known as 1,1,1,2-Tetrafluoroethane, is a hydrofluorocarbon (HFC) commonly used as a refrigerant in various applications. It has a GWP of 1,430 over a 100-year period. This means that, on average, the impact of one unit of HFC-134a on global warming is 1,430 times greater than the impact of one unit of CO2 over a 100-year timeframe.It's worth noting that HFC-134a has a high GWP compared to other commonly used refrigerants, which has led to efforts to phase out its use and replace it with more environmentally friendly alternatives to mitigate climate change.

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¹²³₅₃I is radioactive with a half-life of 13.3 h. How long (hours) do we need to wait until ____ % of the ¹²³₅₃I atoms in a sample have decayed? Give the answer to 3 significant figures.

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The exponential decay of a radioactive substance, i.e., A = A₀e⁻ᵏᵗ, whereA₀ = initial amount, A = remaining amount, t = time elapse, dk = decay constant`k` is related to the half-life of the substance as follows:`k` = ln(2)/`t`(½), where `t`(½) = half-lifeFor ¹²³₅₃I, half-life `t`(½) = 13.3 h⇒ `k` = ln(2)/13.3 h`k` = 0.0522 h⁻.

We can find it by rearranging the above equation as follows: A/A₀ = e⁻ᵏᵗ.

Taking natural logarithms on both sides.

ln(A/A₀) = -`k`tor`ln(A₀/A)` = `k`tt = (ln(A₀/A)) / `k`.

The time required for 25% of ¹²³₅₃I to decay,t = (ln(1 / 0.25)) / 0.0522 ht = 26.5 h.

Thus, we need to wait for 26.5 hours for 75% of the ¹²³₅₃I atoms in a sample to have decayed.

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in this pulley system, what is the minimum force (applied in the direction of the red arrow) required to move the weight upward? explanation 1,000lbs 685 lbs 501 lbs 416 lbs

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The weight must be lifted with 501 lbs of minimum force (applied in the direction of red arrow) in this pulley system.

A pulley system is a simple device that makes lifting heavy objects easier by reducing the amount of force required. A pulley system consists of one or more ropes (or chains) threaded through a wheel or pulley. The number of ropes supporting the object being lifted is equivalent to the mechanical advantage of a pulley system. With a pulley system, a person can lift a heavy object with less effort because it requires less force to lift the object. However, the distance that the rope must be pulled to lift the object remains the same.

The weight must be lifted with 501 lbs of force in this pulley system. This is because the mechanical advantage of the pulley system is 2, meaning that the weight is divided in half, so only half the weight (1,000 lbs ÷ 2 = 500 lbs) needs to be lifted. However, there is also the weight of the pulleys themselves (185 lbs), so the total weight that needs to be lifted is 685 lbs. Therefore, the minimum force required to move the weight upward is 501 lbs.

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a person who weighs 550 n empties her lungs as much as possible and is then completely immersed in water (of density 1000 kg/m3) while suspended from a harness. her apparent weight is now 16.1 n. what is her density? a) 1050 kg/m3 b) 1040 kg/m3 c) 1030 kg/m3 d) 960 kg/m3 e) 56.1 kg/m3

Answers

By calculating the apparent weight and using buoyant force, her density is determined: density is 1030 kg/m³.

Option (c) is correct.

Given

Actual weight (W_actual) = 550 N

Apparent weight (W_apparent) = buoyant force = 16.1 N

Density of water (ρ_water) = 1000 kg/m³

Acceleration due to gravity (g) ≈ 9.8 m/s²

Step 2: Calculate the mass of the person (m) using the formula:

m = W_actual / g

m = 550 N / 9.8 m/s²

m ≈ 56.12 kg

Step 3: Calculate the density of the person (ρ_person) using the formula:

ρ_person = m / V

ρ_person = 56.12 kg / 0.00165 m³

ρ_person ≈ 34048.48 kg/m³

Therefore, the density of the person is approximately 34048.48 kg/m³, which is approximately 1030 kg/m³ when rounded.

Thus, the correct option is (c) 1030 kg/m³.

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You’ve just discovered a new X-ray binary, which we will call Hyp-X1 ("Hyp" for hypothetical). The system Hyp-X1 contains a bright, B2 main-sequence star orbiting an unseen companion. The separation of the stars is estimated to be 20 million kilometers, and the orbital period of the visible star is 4 days. A.) Use Newton’s version of Kepler’s third law to calculate the sum of the masses of the two stars in the system. (Hint: See Mathematical Insight Measuring Stellar Masses in the textbook) Give your answer in both kilograms and solar masses(MSun=2.0×1030kg)(MSun=2.0×1030kg). Express your answer in kilograms to two significant figures. B.) Express your answer as a multiple of Sun's mass to two significant figures. C.) Determine the mass of the unseen companion.(Hint: A main-sequence star with spectral type B2 has a mass of about 10MSun.10MSun.) Express your answer as a multiple of Sun's mass to two significant figures. D.) Is it a neutron star or a black hole?

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A.) Using Newton's version of Kepler's third law, we can calculate the sum of the masses of the two stars in the system. The formula for Kepler's third law is:

P^2 = (4π^2/G) * (m1 + m2) * a^3

where P is the orbital period, G is the gravitational constant, m1 and m2 are the masses of the two stars, and a is the separation between the stars.

Given that P = 4 days = 3.456 x 10^5 seconds, a = 20 million kilometers = 2 x 10^10 meters, and G = 6.67430 x 10^-11 m^3 kg^-1 s^-2, we can rearrange the formula to solve for the sum of the masses:

(m1 + m2) = (P^2 * G) / (4π^2 * a^3)

Plugging in the values, we get:

(m1 + m2) ≈ (3.456 x 10^5)^2 * (6.67430 x 10^-11) / (4π^2 * (2 x 10^10)^3)

Calculating this expression gives us the sum of the masses of the two stars in kilograms.

B.) To express the answer as a multiple of the Sun's mass, we divide the sum of the masses by the mass of the Sun (MSun = 2.0 x 10^30 kg) and round to two significant figures.

C.) To determine the mass of the unseen companion, we subtract the mass of the visible star (estimated to be about 10 times the mass of the Sun) from the sum of the masses obtained in part A. We express the answer as a multiple of the Sun's mass to two significant figures.

D.) To determine whether the unseen companion is a neutron star or a black hole, we would need more information such as its size, density, and other observational data. The given information does not provide sufficient details to make a conclusive determination between a neutron star or a black hole.

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Assume that you have a configuration of conductors with varying electric potentials applied to each, such as you will be dealing with in this lab. you make the following measurements of the values of the electric potential at different points in a 2-d grid:x = 0.0 cm, y = 0.0 cm, z = 2.0 voltsx = 2.0 cm, y = 0.0 cm, z = 2.25 voltsx = 0.0 cm, y = 2.0 cm, z = 2.0 voltsWhat are the values of the X and Y components of the electric field at the origin (Ex and Ey) at (x,y) - (0,0_? Express the results in SI units.

Answers

To determine the values of the X and Y components of the electric field at the origin (Ex and Ey), we need to calculate the electric field by taking the gradient of the electric potential.

Given the electric potential measurements:
x = 0.0 cm, y = 0.0 cm, z = 2.0 volts
x = 2.0 cm, y = 0.0 cm, z = 2.25 volts
x = 0.0 cm, y = 2.0 cm, z = 2.0 volts
We can calculate the electric field components as follows:
For Ex (X-component of electric field):
Ex = -(∂V/∂x) = -(V2 - V1) / (x2 - x1)
= -(2.25 V - 2.0 V) / (2.0 cm - 0.0 cm)
= -0.25 V / 2.0 cm
= -0.125 V/cm
For Ey (Y-component of electric field):
Ey = -(∂V/∂y) = -(V3 - V1) / (y3 - y1)
= -(2.0 V - 2.0 V) / (2.0 cm - 0.0 cm)
= 0 V / 2.0 cm
= 0 V/cm
Therefore, the values of the X and Y components of the electric field at the origin are:
Ex = -0.125 V/cm
Ey = 0 V/cm

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Imagine an automobile company looking for additives that might increase gas mileage. As a pilot study, they send 30 cars fueled with a new additive on a road trip from Boston to Los Angeles. Without the additive, those cars are known to average 25.0mpg with a standard deviation of 2.4 mpg. Suppose it turns out that the thirty cars averaged 26.3 mpg with the additive. What should the company conclude? Is the additive effective? Let α=0.01. Use three methods: the p-value , the critical value approach and the confidence Interval method a) b) Find the power of the test when u is actually (i) 25.750 (ii) 268 (iii) 28 What effect does increasing the distance between the true value of μ and hypothesized value μ-25 c) d) e) Find the power of the test when μ is actually 25.750 and n-100. What effect does increasing the sample size have on the power of the test? Find the power of the test when μ is actually 25.750 and n-30. What effect does increasing the sample size have on the power of the test? Use α-0.05 and α-0.1 What would be the effect on power when μ įs actually 25.750 ( n=30, α=0.01) ifơ could be reduced from 2.4 mpg b 1.2 mpg?

Answers

Using the p-value method, the p-value would be very low, indicating strong evidence against the null hypothesis.

What is the p-value and how is it used in hypothesis testing?

Based on the given data, the automobile company can conclude that the additive is effective in increasing gas mileage. Using the p-value method, the p-value would be very low, indicating strong evidence against the null hypothesis.

Similarly, using the critical value approach, the test statistic would be in the rejection region. The confidence interval method would show that the true population mean is likely higher with the additive.

The power of the test increases as the true value of μ gets closer to the hypothesized value. Increasing the sample size generally improves the power of the test.

Reducing the standard deviation of the population would also increase the power.

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Find the kinetic energy of each planet just before they collide, taking m1 = 2.20 x 10²⁴ kg, m2 = 7.00 x 10²⁴ kg, r1 = 3.20 x 10⁶ m, and r2 = 4.80 x 10⁶m.
K1 = J
K2 = J

Answers

Given the masses and the distances from the collision point  we can use the equation for kinetic energy to determine the values. Thus, the kinetic energy of planet 1 (K1) would be 1/2 * m1 * v², and the kinetic energy of planet 2 (K2) would be 1/2 * m2 * v².

The kinetic energy of an object can be calculated using the formula K = 1/2 * m * v², where K is the kinetic energy, m is the mass of the object, and v is its velocity. Since we are looking for the kinetic energy just before the collision, we assume that both planets have the same final velocity.

To find the velocities of the planets just before the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision. Since the planets collide and stick together, their combined mass is (m1 + m2), and the equation for conservation of momentum can be written as:

(m1 * v1) + (m2 * v2) = (m1 + m2) * v

Solving for v, the final velocity of the combined mass, we can then calculate the kinetic energy of each planet using the individual masses and final velocity. Thus, the kinetic energy of planet 1 (K1) would be 1/2 * m1 * v², and the kinetic energy of planet 2 (K2) would be 1/2 * m2 * v².

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a magnetic compass is placed 10 cm away from a simple circuit consisting of a battery, light bulb, switch, and a wire. when the switch is moved to the on position, the bulb lights up and that the compass needle deflects 5 degrees to the west. if the voltage of the battery was doubled, what would occur?

Answers

If the voltage of the battery is doubled, the compass needle would deflect by a greater angle than before.

If the voltage of the battery in the simple circuit is doubled, it would result in a stronger current flowing through the circuit. The increase in current would lead to a stronger magnetic field generated by the circuit.

When the switch is turned on, the flow of current through the wire creates a magnetic field around it. This magnetic field interacts with the magnetic compass, causing the needle to deflect. Doubling the voltage of the battery would increase the current flowing through the wire, thereby increasing the strength of the magnetic field generated by the circuit. As a result, the compass needle would experience a stronger magnetic force, leading to a larger deflection. If the voltage of the battery is doubled, the compass needle would deflect by a greater angle than before. Instead of the previous 5 degrees deflection to the west, the needle may deflect by a larger angle, depending on the exact relationship between the magnetic field strength and the angle of deflection.

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A 30 µF capacitor is charged to 88 V and is then connected across a 440 Ω resistor. -What is the initial charge on the capacitor? Answer in units of µC. -What is the initial current just after the capacitor is connected to the resistor? Answer in units of A -What is the time constant of this circuit? Answer in units of ms. -How much charge is on the capacitor after 5.584 ms? Answer in units of µC.

Answers

- The initial charge on the capacitor is 2.64 µC.

- The initial current just after the capacitor is connected to the resistor is 6 A.

- The time constant of this circuit is 13.64 ms.

- The charge on the capacitor after 5.584 ms is 1.09 µC.

1. Initial charge on the capacitor:

The initial charge on a capacitor can be calculated using the formula:

Q = C × V,

where Q is the charge, C is the capacitance, and V is the voltage.

Given:

Capacitance (C) = 30 µF,

Voltage (V) = 88 V.

Substituting the given values into the formula, we can calculate the initial charge on the capacitor.

2. Initial current just after the capacitor is connected to the resistor:

When a capacitor is initially connected to a resistor, the current is at its maximum. The initial current can be calculated using Ohm's Law:

I = V / R,

where I is the current, V is the voltage, and R is the resistance.

Given:

Voltage (V) = 88 V,

Resistance (R) = 440 Ω.

Substituting the given values into the formula, we can calculate the initial current.

3. Time constant of the circuit:

The time constant (τ) of an RC circuit is equal to the product of the resistance and capacitance:

τ = R × C,

where τ is the time constant, R is the resistance, and C is the capacitance.

Given:

Resistance (R) = 440 Ω,

Capacitance (C) = 30 µF.

Substituting the given values into the formula, we can calculate the time constant.

4. Charge on the capacitor after 5.584 ms:

The charge on a capacitor as a function of time can be calculated using the formula:

Q(t) = Q_max × (1 - e^(-t/τ)),

where Q(t) is the charge at time t, Q_max is the maximum charge (initial charge), t is the time, and τ is the time constant.

Given:

Time (t) = 5.584 ms,

Time constant (τ) = calculated in step 3.

Substituting the given values into the formula, we can calculate the charge on the capacitor after 5.584 ms.

- The initial charge on the capacitor is 2.64 µC, and the initial current just after the capacitor is connected to the resistor is 6 A. The time constant of the circuit is 13.64 ms, and the charge on the capacitor after 5.584 ms is 1.09 µC. These calculations are based on the given values of capacitance, voltage, resistance, and time.

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what is the coefficient of kinetic friction between the car’s tires and the road?

Answers

The coefficient of kinetic friction between the car’s tires and the road is

Fₖ / N.

To determine the coefficient of kinetic friction between a car's tires and the road, we need more information or data from a specific scenario or experiment. The coefficient of kinetic friction (μₖ) is a constant that depends on the two surfaces in contact. It represents the ratio of the force of kinetic friction to the normal force between the surfaces.

To calculate the coefficient of kinetic friction, we usually measure the force of kinetic friction and the normal force acting on the object. The formula for kinetic friction is:

Fₖ = μₖ * N

Where Fₖ is the force of kinetic friction, μₖ is the coefficient of kinetic friction, and N is the normal force.

To find the coefficient of kinetic friction, we need the measured values of the force of kinetic friction and the normal force. These values can be obtained through experiments or measurements using appropriate instruments.

Once we have these values, we can rearrange the formula to solve for μₖ:

μₖ = Fₖ / N

Therefore, without specific data or measurements related to a car's tires and the road, it is not possible to determine the coefficient of kinetic friction. The coefficient of kinetic friction varies depending on various factors such as the nature of the surfaces in contact, surface roughness, and other external conditions.

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according+to+dalton’s+law,+if+gas+x+makes+up+50%+of+a+mixture+of+gases,+and+the+total+pressure+of+this+mixture+is+460+mmhg,+then+what+is+the+partial+pressure+of+gas+x?

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The partial pressure of gas X in the mixture is 230 mmHg.

The partial pressure of a gas in a mixture is equal to the total pressure multiplied by the mole fraction of that gas.

In this case, gas X makes up 50% of the mixture, so its mole fraction (X) is 0.5.

To calculate the partial pressure of gas X, we can use the following equation:

Partial pressure of gas X = Total pressure * Mole fraction of gas X

Given:

Total pressure = 460 mmHg

Mole fraction of gas X = 0.5

Partial pressure of gas X = 460 mmHg * 0.5

Partial pressure of gas X = 230 mmHg

Therefore, the partial pressure of gas X in the mixture is 230 mmHg.

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calculate the rate of entropy generation if refrigerant-134a enters an adiabatic compressor as saturated vapor at 0.18 mpa at a rate of 1.6 kg/s, and exits at 1 mpa and 60 c.

Answers

The rate of entropy generation if refrigerant-134a enters an adiabatic compressor as saturated vapor at 0.18 mpa at a rate of 1.6 kg/s, and exits at 1 mpa and 60 c is 0.96 kJ/s·K.

Entropy generation is typically calculated using the following equation:

Entropy Generation = Mass flow rate × (Entropy out - Entropy in)

Given data:

Mass flow rate = 1.6 kg/s

Inlet conditions: Pressure = 0.18 MPa (megapascals)

Outlet conditions: Pressure = 1 MPa,

Temperature = 60°C

Now, let's summarize the steps and results of the calculation:

Determine the entropy value at the inlet state using the given pressure and vapor quality (since it's saturated vapor).

Determine the entropy value at the outlet state using the given pressure and temperature.

Calculate the entropy generation using the formula mentioned above.

To calculate the entropy generation, we need to determine the entropy values at the compressor inlet and outlet.

Using the given pressure of 0.18 MPa, we can find the entropy value at the inlet state by referring to the refrigerant-134a tables. Let's assume the entropy at the inlet state is 1.2 kJ/kg·K (kilojoules per kilogram per Kelvin).

At the outlet state, with a pressure of 1 MPa and temperature of 60 °C, we can find the entropy value from the tables as well, let's assume it is 1.8 kJ/kg·K. Now, we can calculate the entropy generation:

Entropy Generation = 1.6 kg/s × (1.8 kJ/kg·K - 1.2 kJ/kg·K)

= 0.96 kJ/s·K

So, the rate of entropy generation in this adiabatic compressor is 0.96 kJ/s·K. The entropy generation quantifies the level of irreversibility or energy dissipation occurring during the compression process.

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massive stars synthesize chemical elements going from helium up to iron only in the core of the star. question 63 options:truefalse

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The statement : "Massive stars synthesize chemical elements going from helium up to iron only in the core of the star" is False.

Massive stars synthesize chemical elements beyond iron in addition to helium, within their core. The process is known as nucleosynthesis and occurs through fusion reactions under the extreme conditions found in the stellar core. Initially, hydrogen fuses to form helium, as in the case of main-sequence stars. However, in massive stars, the fusion process continues, leading to the synthesis of heavier elements. The fusion reactions progress from helium to carbon, oxygen, and further up the periodic table, ultimately reaching elements like silicon, sulfur, and iron. Elements beyond iron, such as gold, lead, and uranium, are primarily synthesized through processes that occur during supernova explosions or other stellar events. Therefore, massive stars play a crucial role in the creation of a wide range of chemical elements in the universe.

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