Question 8 A 2.7 m long string vibrates as a three loop standing wave. The amplitude is 0.98 cm and wave speed is 90 m/s. Part A Find the frequency of the vibration. μA ? f= Value Submit Provide Feed

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Answer 1

The frequency of the vibration is approximately 50 Hz, calculated using the formula v = fλ, where v is the wave speed and λ is the wavelength determined by the length of the string and the number of loops.

In a standing wave, the length of the string can be related to the wavelength of the wave by the equation:

λ = 2L/n,

where λ is the wavelength, L is the length of the string, and n is the number of loops.

In this case, the length of the string L is given as 2.7 m, and the number of loops n is 3. Plugging in these values, we can solve for the wavelength:

λ = 2(2.7 m)/3

λ = 1.8 m.

The wave speed v is given as 90 m/s. The frequency f can be calculated using the formula:

v = fλ.

Rearranging the equation, we have:

f = v/λ = 90 m/s / 1.8 m

f = 50 Hz.

Therefore, the frequency of the vibration is approximately 50 Hz.

The frequency of the vibration is approximately 50 Hz, calculated using the formula v = fλ, where v is the wave speed and λ is the wavelength determined by the length of the string and the number of loops.

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Related Questions

three swimmers who all swim at the same speed discuss how to cross a river in the shortest amount of time. swimmer a will swim straight across the river at a right angle to the current. swimmer b reasons that the current will carry a downstream, meaning that a will cover a greater distance to get across and therefore will take a longer time interval. b says he will aim at an upstream angle such that, allowing for the current, he will reach the other side directly across from where he starts, thus covering the shortest distance and arriving first. swimmer c , reasoning that the time interval needed for b to cross will be longer than b expects because some of b 's effort will be spent battling the current, plans to aim at a downstream angle, so that the current assists rather than opposes him. this way he will be traveling at the highest speed and get across first. part a which swimmer gets across first? which swimmer gets across first? swimmer a gets across first. all swimmers get across at the same time. swimmer b gets across first. swimmer c gets across first. request answer provide feedback

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Swimmer C gets across first. The reasoning behind the three swimmers crossing the river is different. Swimmer A swims directly across the river. Swimmer B swims upstream at an angle that allows him to be carried downstream by the current and reach the other side directly across from his starting point.

Swimmer C swims downstream at an angle that allows him to be carried downstream by the current, therefore, making the current work for him. Therefore, Swimmer C gets across first because he is swimming at the highest speed due to the help of the current. Swimmer B will take longer than he expects because some of his efforts will be spent battling the current. Swimmer A will cover the longest distance and will take longer than both Swimmer B and Swimmer C. Hence, Swimmer C is the one who gets across first.

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2) (5 points) What is the percent uncertainty of the measurement (3.654 ± 0.1) km? Make sure it is properly rounded. 3) (5 points) If v = Av² t + B √ + Ct².1 + Ct². Using dimensional analysis ob

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a) The percent uncertainty of a measurement can be calculated using the formula:

Percent uncertainty = (Uncertainty / Measurement) * 100

In this case, the measurement is (3.654 ± 0.1) km. The uncertainty is ±0.1 km. Therefore, the percent uncertainty is:

Percent uncertainty = (0.1 km / 3.654 km) * 100 = 2.74%

b) The given equation is:

v = Av² t + B √ + Ct².1 + Ct²

Using dimensional analysis, we can analyze the dimensions of each term in the equation to determine the dimensions of the variables A, B, and C.

The dimensions of the left-hand side (v) are [L]/[T], representing velocity.

Analyzing each term on the right-hand side:

- Av² t has dimensions of [L²]/[T²] * [T] = [L²]/[T]

- B √ has dimensions of [L]

- Ct².1 has dimensions of [L]/[T²] * [T².1] = [L]

To have consistent dimensions on both sides of the equation, A must have dimensions of [1]/[T], B must have dimensions of [L], and C must have dimensions of [1]/[T²].

Dimensional analysis allows us to check the correctness of equations and identify the dimensions of unknown variables based on the known dimensions of other terms.

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in the upper atmosphere at altitudes where commercial airlines travel, we find extremly cold temperatures what is the speed of sound for a temperature of -49 degrees c

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The speed of sound in dry air at -49°C is approximately 294 meters per second (m/s).

In the upper atmosphere, the temperature drops below -49°C, which is very cold. At these altitudes, commercial airlines fly. The speed of sound in a medium, such as air, is dependent on the temperature of that medium. As a result, the speed of sound in the upper atmosphere at -49°C is different from the speed of sound at room temperature.

The speed of sound is determined by the medium it travels through, as mentioned earlier. The speed of sound in dry air at room temperature is approximately 343 meters per second (m/s). The speed of sound is calculated by the following formula:

Speed of sound = √(γ × R × T), where γ is the ratio of specific heat capacities, R is the gas constant, and T is the temperature in Kelvin.

At the temperature of -49°C, the speed of sound is slower than at room temperature due to the change in temperature. The sound speed decreases with temperature because air molecules are more tightly packed at lower temperatures, causing sound waves to move slower. The speed of sound in dry air at -49°C is approximately 294 meters per second (m/s). This is around 15% slower than the speed of sound at room temperature. As a result, the aircraft should fly at a lower speed than it would at room temperature to compensate for the slower speed of sound at that altitude. Because the speed of sound is slower at colder temperatures, aircraft pilots must be aware of this and account for it when flying in the upper atmosphere. A pilot who is unaware of the change in sound speed could overestimate their speed and fly too fast. This might be harmful to the aircraft and its passengers.

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Assume a 4800 nT/min geomagnetic storm disturbance hit the United States. You are tasked with estimating the economic damage resulting from the storm. a. If there were no power outages, how much impact (in dollars) would there be in the United States just from the "value of lost load?" Explain the assumptions you are making in your estimate. [ If you are stuck, you can assume 200 GW of lost load for 10 hours and a "value of lost load" of $7,500 per MWh.] b. If two large power grids collapse and 130 million people are without power for 2 months, how much economic impact would that cause to the United States? Explain the assumptions you are making in your estimate. c. If every country above 40° magnetic latitude had a similar proportion of its economy impacted, what would the global economic impact (in dollars) be?

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The economic damage estimation resulting from a 4800 nT/min geomagnetic storm disturbance hitting the United States involves assessing the "value of lost load" and the impact of power outages.

a. If there were no power outages, the economic impact in the United States would be determined by the "value of the lost load." Assuming 200 GW of lost load for 10 hours and a value of the lost load at $7,500 per MWh, the calculation would involve multiplying the lost load (in MWh) by the value of the lost load to obtain the dollar amount.

b. If two large power grids collapse and 130 million people experience a 2-month power outage, the economic impact would be significant. However, the estimation would require further assumptions such as the average electricity consumption per person, the GDP loss per day, and the cost of recovery efforts.

c. To determine the global economic impact, a similar proportion of the economy impacted would be considered for every country above the 40° magnetic latitude. Assumptions regarding the affected countries, their respective economies, and the proportion of impact would need to be made to calculate the overall global economic impact.

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Complete the following statements regarding locations and functions of cranial meninges by typing in the correct answer: The and the meningeal layer together compose the dura mater in the cranial cavity. The subarachnoid space contains a protective The , a dural septum, is located within the longitudinal fissure between the cerebral hemispheres. The superior sagittai sinus collects and contains The delicate membrane is located on the surface of the brain.

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The periosteal layer and the meningeal layer together compose the dura mater in the cranial cavity. The subarachnoid space contains a protective cerebrospinal fluid (CSF). The falx cerebri, a dural septum, is located within the longitudinal fissure between the cerebral hemispheres. The superior sagittal sinus collects and contains venous blood. The delicate membrane is located on the surface of the brain.

The dura mater, one of the cranial meninges, is composed of two layers: the periosteal layer, which is attached to the inner surface of the skull, and the meningeal layer, which is deeper and forms a protective covering around the brain. The subarachnoid space is a region filled with cerebrospinal fluid (CSF) that surrounds the brain and spinal cord, acting as a cushion and providing protection. The falx cerebri is a dural septum that runs within the longitudinal fissure, separating the two cerebral hemispheres. It helps to stabilize and support the brain's structures. The superior sagittal sinus is a large venous channel located within the falx cerebri. It collects deoxygenated blood from the brain and carries it back towards the heart. On the surface of the brain, there is a delicate membrane known as the arachnoid mater. It lies between the dura mater and the pia mater and plays a role in protecting the underlying brain tissue. Together, these structures and spaces form part of the complex system of cranial meninges, providing protection, support, and fluid-filled spaces within the cranial cavity.

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Which of the following is not a correct, relevant statement regarding Kepler's 2nd Law?
a. The planet speeds up in its orbit when closer to the Sun, and slows down in its orbit when farther from the Sun
b. The area of the angle "swept out" by an imaginary line conneting the planet and the Sun, during some fixed amount of time, never changes during its orbit around the Sun
c. The planet speeds up in its rotation about its axis when closer to the Sun, and slows down in its rotation about its axis when farther from the Sun
d. The planet conserves angular momentum during its orbit around the Sun

Answers

"The planet speeds up in its rotation about its axis when closer to the Sun, and slows down in its rotation about its axis when farther from the Sun" is not a correct, relevant statement regarding Kepler's 2nd Law.

The correct answer is option C.

Kepler's 2nd Law, also known as the Law of Equal Areas, states that a line connecting a planet to the Sun sweeps out equal areas in equal times as the planet moves in its elliptical orbit around the Sun. This law is relevant to the orbital motion of the planet, not its rotation about its axis.

The rotation of a planet about its axis is governed by other factors, such as its own internal forces and torques. The distance from the Sun does not directly affect the planet's rotation about its axis. Therefore, statement c is not a correct or relevant statement regarding Kepler's 2nd Law.

The correct statements regarding Kepler's 2nd Law are:

a. The planet speeds up in its orbit when closer to the Sun and slows down in its orbit when farther from the Sun. This is because the planet experiences a stronger gravitational force from the Sun when it is closer, resulting in a higher orbital speed.

b. The area of the angle "swept out" by an imaginary line connecting the planet and the Sun, during some fixed amount of time, never changes during its orbit around the Sun. This implies that the planet covers equal areas in equal times, reflecting the conservation of angular momentum in the absence of external torques.

d. The planet conserves angular momentum during its orbit around the Sun. This means that the product of the planet's moment of inertia and its angular velocity remains constant throughout its orbit, in the absence of external torques.

In summary, statement c is not a correct or relevant statement regarding Kepler's 2nd Law, while statements a, b, and d accurately describe the key aspects of Kepler's 2nd Law.

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As a light ray enters or exits a water-air interface at an angle of 15 degrees with the normal, it

-always bends away from the normal
-sometimes bends away from the normal
-always bends towards the normal
-does not bend

Answers

When a light ray enters or exits a water-air interface at an angle of 15 degrees with the normal, it will bend towards the normal. \

This phenomenon is known as refraction. Refraction occurs because light travels at different speeds in different mediums. In this case, light travels slower in water than in air. According to Snell’s law, the angle of refraction is related to the angle of incidence and the refractive indices of the two mediums. The refractive index of water is higher than that of air, which means that light rays will bend towards the normal when entering water. Since the incident angle of 15 degrees is less than the critical angle, total internal reflection does not occur. Instead, the light ray will bend towards the normal as it enters the water-air interface. When the light ray exits the interface, it will once again bend towards the normal. This bending of light is responsible for various optical phenomena, such as the apparent bending of a straight object when partially immersed in water (refraction in a glass of water) and the formation of rainbows. Understanding the principles of refraction is crucial in fields such as optics, physics, and engineering, as it governs the behavior of light at interfaces between different mediums.

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what wavelength of radiation has photons of energy 6.06 x 10^-19 j

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The wavelength of radiation that has photons of energy 6.06 x 10^-19 J is 3.288 × 10⁻⁷ m.

The formula that can be used to solve the given question is given as;

                                        E = hc/λ                            where E = energy of the photon, h = Planck's constant, c = speed of light, λ = wavelength of light.

                            h = 6.626 × 10⁻³⁴ Js, c = 3 × 10⁸ m/s, E = 6.06 × 10⁻¹⁹ J

Substitute all the values into the equation

                                         E = hc/λ;

                                    6.06 × 10⁻¹⁹

                                   = 6.626 × 10⁻³⁴ × 3 × 10⁸ / λRearrange for λ;

                                   λ = hc/E = (6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s) / 6.06 × 10⁻¹⁹ J

                                          = 3.288 × 10⁻⁷ m

Therefore, the wavelength of radiation that has photons of energy 6.06 x 10^-19 J is 3.288 × 10⁻⁷ m

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if a person pulls on a cart to the right with a force of 10n and a second person pulls to the left with a force of 3n, what is the net force direction on the cart?

Answers

The net direction on the cart, if a person pulls on a cart to the right with a force of 10n and a second person pulls to the left with a force of 3n, is 7n to the right.

The net force direction can be determined by finding the net vector sum of the forces acting on it.

Since the first person pulled the cart to the right with a force of 10n,

Assuming the right direction as positive, this force can be represented as a vector ⇒(+)10n

Similarly, as the second person pulls the cart to the left, the force can be represented as a vector ⇒ (-)3n,

Net force = Net vector sum of the forces

                   =(+10n) + (-3n)

Net force = +7n

Thus, the net direction on the cart, if a person pulls on a cart to the right with a force of 10n and a second person pulls to the left with a force of 3n, is 7n to the right.

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Answer: 7n

Explanation:

Determine how long (in years) it would take for you to reach (a) the start Proxima Centauri at a distance of 4.3 light years away. Assume you are traveling in the fastest spacecraft ever constructed by mankind (approx 150,000 mph), (b) Repeat for a journey to the center of our Milky Way at a distance of 25,000 light years away. Show all of your work.

Answers

Answer:

Journey to Proxima Centauri:

150,000 miles/hour * 24 hours/day * 365 days/year = 1,314,000,000 miles/year

Travel time = Distance / Speed

Travel time = 4.3 light years / (1,314,000,000 miles/year)

Travel time ≈ 3.273 years

It would take approximately 3.273 years to reach Proxima Centauri.

Journey to the center of the Milky Way:

Distance to the center of the Milky Way: 25,000 light years

Speed of the spacecraft: 150,000 miles per hour

Travel time = Distance / Speed

Travel time = 25,000 light years / (1,314,000,000 miles/year)

Travel time ≈ 19019.14 years

It would take approximately 19,019.14 years to reach the center of the Milky Way.

as waves approach the shore, do their heights increase or decrease? do wavelengths become longer or shorter?

Answers

As waves approach the shore, their heights increase while their wavelengths decrease, which is also known as shoaling.

As the wave approaches the shore, the lower portion of the wave touches the seabed and slows down, while the top continues at its original speed. This results in a reduction in wavelength and an increase in wave height.

This is due to the conservation of energy principle, which states that energy can neither be created nor destroyed.

As a result, the energy in the wave is compressed into a smaller space as it approaches the shore, resulting in an increase in wave height.

As the wave approaches the shore, the lower portion of the wave touches the seabed and slows down, while the top continues at its original speed. This results in a reduction in wavelength and an increase in wave height.

This is due to the conservation of energy principle, which states that energy can neither be created nor destroyed.

As a result, the energy in the wave is compressed into a smaller space as it approaches the shore, resulting in an increase in wave height.

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A superconducting wire carries a current of 1000 A with a radius
of 0.8 m. Find the magnetic field (in ×10-4 T) at the
radius of the wire.

Answers

The magnetic field at the radius of the superconducting wire is 25 × 10^(-4) T.

To calculate the magnetic field at the radius of the superconducting wire carrying a current, you can use Ampere's law, which relates the magnetic field to the current enclosed by a closed loop around the wire.

The formula for the magnetic field inside a wire is given by where B is the magnetic field, μ₀ is the permeability of free space (4π × 10^(-7) T m/A), I is the current, and r is the radius of the wire.

Plugging in the given values:

I = 1000 A (current),

r = 0.8 m (radius), and

μ₀ = 4π × 10^(-7) T m/A (permeability of free space),

B = (4π × 10^(-7) T m/A * 1000 A) / (2 * π * 0.8 m).

Simplifying the expression:

B = (4 * 10^(-7) T m) / (1.6 m).

B = 2.5 * 10^(-7) T.

Finally, to convert the magnetic field to the requested format (×10^(-4) T), we can express 2.5 * 10^(-7) T as 25 * 10^(-9) T or 25 × 10^(-4) T.

Therefore, the magnetic field at the radius of the superconducting wire is 25 × 10^(-4) T.

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What does it mean to say that energy is conserved?

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Energy cannot be created or destroyed meaning the total amount of energy never changes.

The total energy of a system remains constant in an isolated system, according to the law of conservation of energy.

Energy can be changed from one form to another, for as converting potential energy to kinetic energy, but the overall amount of energy in the domain never changes.

The kinetic energy that is lost as a body slows down when ascending against the pull of gravity was thought to be turned into potential energy, or stored energy, which was then converted back into kinetic energy when the body accelerated upon returning to Earth.

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A 0.420-m-long guitar string, of cross-sectional area 1.00 x 10-6 m², has Young's modulus Y= 2.00 GPa. By how much must you stretch the string to obtain a tension of 15.1 N? mm

Answers

We would need to stretch the guitar string by approximately 3.15 mm to obtain a tension of 15.1 N.

To calculate the amount by which the guitar string must be stretched to obtain a tension of 15.1 N, we can use Hooke's Law, which states that the tension in a stretched string is directly proportional to the extension or change in length.

The formula for Hooke's Law is:

T = Y * A * (ΔL / L),

where:

T is the tension in the string,

Y is the Young's modulus of the material,

A is the cross-sectional area of the string,

ΔL is the change in length or extension of the string, and

L is the original length of the string.

We are given the following values:

L = 0.420 m (original length of the string),

A = 1.00 × 10⁻⁶ m² (cross-sectional area of the string),

Y = 2.00 GPa = 2.00 × 10⁹ Pa (Young's modulus of the string), and

T = 15.1 N (desired tension in the string).

Let's substitute these values into the formula and solve for ΔL:

15.1 N = (2.00 × 10⁹Pa) * (1.00 × 10⁻⁶ m²) * (ΔL / 0.420 m).

To solve for ΔL, we rearrange the equation:

ΔL = (15.1 N * 0.420 m) / (2.00 × 10⁹Pa * 1.00 × 10⁻⁶ m²).

Simplifying the equation:

ΔL = 3.15 × 10⁻³ m.

To convert ΔL to millimeters (mm), we multiply by 1000:

ΔL = 3.15 × 10⁻³ m * 1000

ΔL = 3.15 mm.

Therefore, you would need to stretch the guitar string by approximately 3.15 mm to obtain a tension of 15.1 N.

In conclusion, to achieve a tension of 15.1 N in a 0.420-m-long guitar string with a cross-sectional area of 1.00 × 10⁻⁶ m² and a Young's modulus of 2.00 GPa, you would need to stretch the string by approximately 3.15 mm.

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Guide Questions:
1. Vertical displacement is called
2. Horizontal displacement is called
3. What is the horizontal velocity at 3 sec?
4. What is the horizontal velocity at 4 sec?
5. At what time did the ball reach the maximum height?
6. What is the horizontal velocity at 5 sec?
7. What is the horizontal velocity at 6 sec, when the ball is going down?
8. Describe the horizontal velocity as the ball goes up.
9. Describe the horizontal velocity as the ball goes down.​

Answers

Vertical displacement is called "vertical distance", Horizontal displacement is called "horizontal distance",  the horizontal velocity is 6 m/s at 3 seconds, the horizontal velocity at 4 sec is 6m/sec, the ball reach the maximum height at 2 sec,  the horizontal velocity remains constant when the ball goes up and down.

1. Vertical displacement is called "vertical distance" or "vertical displacement" and refers to the change in height or position along the vertical axis.

2. Horizontal displacement is called "horizontal distance" or "horizontal displacement" and refers to the change in position along the horizontal axis.

3. To determine the horizontal velocity at 3 seconds, we can use the information provided in the image you shared. The horizontal velocity remains constant throughout the motion. Looking at the graph, we can see that the horizontal velocity is 6 m/s at 3 seconds, as indicated by the constant horizontal line.

4. Similar to the previous question, we can determine the horizontal velocity at 4 seconds from the graph. It appears to be 6 m/s at 4 seconds.

5. The time at which the ball reaches its maximum height can be determined by finding the highest point on the vertical displacement graph. In the graph you shared, the maximum height is reached at around 2 seconds.

6. Based on the information in the graph, it is not possible to determine the horizontal velocity at 5 seconds. There is no corresponding data point or line indicating the velocity at that specific time.

7. Similarly, the graph does not provide information about the horizontal velocity at 6 seconds when the ball is going down. We cannot determine it from the given data.

8. As the ball goes up, the horizontal velocity remains constant. From the graph, we can observe a horizontal line indicating a constant horizontal velocity during the upward motion.

9. As the ball goes down, the horizontal velocity also remains constant. The graph suggests that the horizontal velocity remains the same during the downward motion, as indicated by the horizontal line.

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DETAILS SERCP11 24.4.OP.019. An extremely thin sheet of glass is being inspected at the camera store. Illuminated by white light at near-normal incidence, the f refractive index of 1.52, what wavelength of visible light (in nm) does it reflect most strongly? (The wavelengths of visible light ras nm Need Help? Read It mera store. Illuminated by white light at near-normal incidence, the film-like sheet is 0.384 µm thick and has air on both sides. If the glass has a m) does it reflect most strongly? (The wavelengths of visible light range from 400 to 700 nm.)

Answers

The wavelength of visible light that an extremely thin sheet of glass reflects most strongly is 1150 nm.

In order to find out the wavelength of visible light that an extremely thin sheet of glass reflects most strongly, we need to use the following formula:

2nt=mλ

Where:

n is the refractive index

t is the thickness

m is the order of the reflected wavelength

λ is the wavelength of light

First, we have to determine which order of reflected light will be the strongest, then solve for λ by dividing by m.

1.52 × 0.385 × 10^-6 m = mλ

If we assume that the reflection comes from the first order (m = 1), then,

λ = 2nt/m= 2(1.52)(0.385 × 10^-6)/(1)

≈1.15 × 10^-6 m

= 1150 nm (rounded to the nearest nm).

Therefore, the wavelength of visible light in nm that the sheet of glass reflects most strongly is 1150 nm.

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Part E For both Tracker experiments, calculate the average vertical velocity, where the time period is t = 0.00 second to t = 1.00 second. Consider only the magnitude of the displacement. Record your results to three significant figures. Comment: Which ball drops faster during the first second of the fall?

Answers

Answer:To calculate the average vertical velocity for both Tracker experiments, we need the magnitude of the displacement and the time period from t = 0.00 seconds to t = 1.00 second. However, you have mentioned "Part E" and "Which ball drops faster," indicating that there is a previous context or specific experiment being referred to that I'm unaware of.

Please provide the necessary information or context regarding the Tracker experiments, such as the initial heights or any other relevant details, so that I can assist you with calculating the average vertical velocity and determining which ball drops faster.

Explanation:

To calculate the average vertical velocity for the time period between t = 0.00 s and t = 1.00 s, considering only the magnitude of the displacement, we can use the following formula:

Average vertical velocity = Magnitude of displacement / Time interval

For the first ball, we have:

Average vertical velocity = 2.70 m / 1.00 s = 2.70 m/s

For the second ball, we have:

Average vertical velocity = 2.75 m / 1.00 s = 2.75 m/s

Therefore, the second ball drops faster during the first second of the fall, as it has a higher average vertical velocity than the first ball. This result is consistent with the previous analysis where we considered the average vertical acceleration and the magnitude of the displacement separately.

a 2.00-m long steinway piano string of mass 10.0g is under a tension of 320 N. Find the speed with which a wave travels on this string.

Answers

the speed with which a wave travels on this Steinway piano string is 400 m/s.

Explanation:

The speed of a wave on a string is given by the equation:

v = sqrt(T/μ)

where v is the speed of the wave, T is the tension in the string, and μ is the linear mass density of the string.

The linear mass density μ of the string is given by the mass per unit length, which in this case is:

μ = m/L

where m is the mass of the string, and L is the length of the string.

Substituting the given values, we have:

μ = m/L = 10.0 g / 2.00 m = 5.00 g/m = 0.0050 kg/m

Now, using the equation above, we have:

v = sqrt(T/μ) = sqrt(320 N / 0.0050 kg/m) = 400 m/s

Therefore, the speed with which a wave travels on this Steinway piano string is 400 m/s.

The speed of a wave traveling on a 2.00-meter long Steinway piano string, which has a mass of 10.0 grams and is under a tension of 320 N, needs to be determined.

The speed of a wave traveling on a string can be calculated using the formula [tex]v = \sqrt[] (T/\mu)[/tex], where v is the wave speed, T is the tension in the string, and μ is the linear mass density of the string. The linear mass density is given by the equation μ = m/L, where m is the mass of the string and L is its length.

In this case, the mass of the string is 10.0 grams (or 0.010 kg) and its length is 2.00 meters. Therefore, the linear mass density μ = 0.010 kg / 2.00 m = 0.005 kg/m.

Now, substituting the values into the wave speed formula, we have[tex]v = \sqrt[](320 N / 0.005 kg/m) = \sqrt[](64000 m^2/s^2 / kg/m) = \sqrt[] (64000 m/s^2) = 253.55 m/s.[/tex]

Therefore, the speed with which the wave travels on the 2.00-meter long Steinway piano string is approximately 253.55 m/s.

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Consider the solutions of the following equation over the interval 0 to 27 . or the interval 0° to 360°. Of the choices shown, which is not a solution to the equation? sec²0 tan0 = 2 tan O 225 degr

Answers

The equation sec²0 tan0 = 2 tan0 can be simplified to tan²0 = 2. The solutions to tan²0 = 2 are 0, 45, 135, and 225 degrees. The choices that are not solutions are 30, 180, and 315 degrees.

The equation sec²0 tan0 = 2 tan0 can be simplified to tan²0 = 2.

The solutions to tan²0 = 2 are 0 degrees, 45 degrees, 135 degrees, and 225 degrees.

Therefore, the answer is 30 degrees, 180 degrees, and 315 degrees.

Here is a more detailed explanation of how to solve the equation:

1. First, we need to simplify the equation. We can do this by using the identity sec²θ = 1 + tan²θ. This gives us the equation tan²θ = 2 - 2tanθ.

2. Now, we can factor the left-hand side of the equation. This gives us (tanθ - 1)(tanθ + 2) = 0.

3. This means that either tanθ = 1 or tanθ = -2.

4. The solutions to tanθ = 1 are 0 degrees and 45 degrees.

5. The solutions to tanθ = -2 are 135 degrees and 225 degrees.

Therefore, the solutions to the original equation are 0 degrees, 45 degrees, 135 degrees, and 225 degrees.

The choices that are not solutions are 30 degrees, 180 degrees, and 315 degrees. This is because tan30 = [tex]\frac{\sqrt{3}}{3}[/tex], tan180 = 0, and tan315 = [tex]-\frac{\sqrt{3}}{3}[/tex]. None of these values are equal to 1 or -2.

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Complete question :

Consider the solutions of the following equation over the interval 0 to 27 . or the interval 0° to 360°. Of the choices shown, which is not a solution to the equation? sec²0 tan0 = 2 tan O 225 degrees O 0 degrees All of the choices shown are solutions. O The answer is not among the choices shown. 0 All of the choices shown are not solutions. 07 135 degrees O 30 degrees O 180 degrees OT O 315 degrees A

Write the expression as the sine, cosine, or tangent of an angle. Then find the exact value of the expression. sin 20° cos 40° + cos 20° sin 40° GELEN Write the expression as the sine, cosine, or

Answers

The expression sin 20° cos 40° + cos 20° sin 40°. So, the exact value of the given expression is √3/2.

The expression sin 20° cos 40° + cos 20° sin 40° can be written using the trigonometric identity for the sine of the sum of two angles:

sin(A + B) = sin A cos B + cos A sin B

Comparing this with the given expression, we can see that it matches the form of the sine of the sum of two angles. Therefore, we can rewrite the expression as:

sin(20° + 40°)

Now, let's calculate the exact value of the expression:

sin(20° + 40°) = sin 60°

The exact value of sine 60° is √3/2.

Therefore, the exact value of the given expression is √3/2.

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complete question :

Write the expression as the sine, cosine, or tangent of an angle. Then find the exact value of the expression.

sin 20° cos 40° + cos 20° sin 40° = ?

What is the average velocity of a bus that moves 38.0 m across a
lake in 3 mins?
Group of answer choices
1.02 m/s across the lake
1.02 m/s North
0.21 m/s North
0.21 m/s across the lake

Answers

The correct answer is option D 0.21m/s across the lake.

Average velocity is defined as the displacement of an object divided by the time taken to cover that displacement. Mathematically, average velocity (avg velocity) can be calculated as:

avg velocity= Δx / Δt

Where:

Δx represents the change in position or displacement of the object,

Δt represents the change in time. To find the average velocity of a bus that moves 38.0 m across a lake in 3 mins, we

need to convert minutes into  seconds. We can then use the formula for velocity to solve for the answer. The formula

for velocity is given as: Velocity  = distance / time. Therefore, Velocity = 38.0 m / (3 x 60 seconds)Velocity = 38.0 m /

180 seconds Velocity = 0.21 m/s across the lake. Hence, the average velocity of a bus that moves 38.0 m across a lake  

in 3 mins is 0.21 m/s across the lake. Therefore, the correct answer is option D.

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Two equally charged identical small balls kept some fixed distance apart exert a repulsive force F on each other. A similar uncharged ball, after touching one of them is placed at the mid-point of line joining the two balls. Force experienced by the third ball is :

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Two equally charged identical small balls kept some fixed distance apart exert a repulsive force F on each other. A similar uncharged ball, after touching one of them is placed at the mid-point of line joining the two balls. Force experienced by the third ball is is zero.

The force experienced by the third ball after it is placed at the midpoint of the line joining the two equally charged balls can be determined using the principle of superposition. Initially, when the two equally charged identical small balls are kept a fixed distance apart, they exert a repulsive force F on each other. Let's call this force F1. When the uncharged ball touches one of the charged balls, it acquires the same charge due to the process of conduction. Now, there are two charged balls with equal and opposite charges, and the uncharged ball with the same charge in between them. Due to the principle of superposition, the force experienced by the third ball is the vector sum of the forces exerted by the two charged balls individually. Let's call the force experienced by the third ball as F3. Since the charged balls have equal and opposite charges, the magnitude of the force exerted by each charged ball on the third ball will be equal and their directions will be opposite. Therefore, the magnitudes of the forces cancel each other out, and the net force experienced by the third ball is zero. Hence, the force experienced by the third ball is zero when it is placed at the midpoint of the line joining the two equally charged balls.

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a box is being pulled to the right. the free body diagram is shown. what is the magnitude of the kinetic frictional force? 25 n 125 n 375 n 500 n

Answers

The magnitude of the kinetic frictional force acting on the box is approximately 9.8 N.

To calculate the magnitude of the kinetic frictional force, we can use the formula:

Frictional force = coefficient of kinetic friction * normal force,

where the normal force is equal to the weight of the box.

The weight of the box can be calculated using the formula:

Weight = mass * gravitational acceleration,

where the gravitational acceleration is approximately [tex]9.8 m/s^2[/tex].

Substituting the given values, we have:

Weight = 5 kg * [tex]9.8 m/s^2[/tex] = 49 N.

Now we can calculate the magnitude of the kinetic frictional force:

Frictional force = 0.2 * 49 N = 9.8 N.

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--The complete Question is, A box is being pulled to the right with a force of 100 N. The box experiences a kinetic frictional force. The mass of the box is 5 kg and the coefficient of kinetic friction between the box and the surface is 0.2. What is the magnitude of the kinetic frictional force acting on the box? --

Answer the following question about nature of galaxies.
Explain how astronomers determined that there are galaxies other than the Milky Way.
[2 marks]
Give three reasons (with explanations) why astronomers were confused about whether there were galaxies other than our own. [3 marks]
Describe the Hubble "tuning fork" diagram, and what it tells us. [3 marks]
Describe the large scale structure of the universe. [2 marks]

Answers

The discovery of other galaxies other than Milky Way is one of the remarkable achievement and discoveries of Astronomy.

How the astronomers determined other galaxies other than Milky Way

There was observational limitationsThe great debateThe discovery of Cepheid variablesThere observational measurementThe red shift and expanding universeSystematic surveyDeep imaging and Hubble space telescope

Reasons for confusion about other galaxies other than Milky Way

Limitation of observational technology.. The telescope used then was not as sophisticated as the later ones. The procurement of new technologies afford the ability to research more and led to doubting of existence of other galaxies.

Misinterpretation of Nebulae. Nebulae are vast clouds of gas and dust in space that can be visually striking. Some nebula are misinterpreted as part of Milky Way.

The Island Universe debate. This debate emerged in the early 20th century regarding the nature of spiral nebulae. Some astronomers argued that that spiral nebulae were smaller "island universes" similar to our Milky Way. Some other scientists believed that argued that these nebulae were nearby gas clouds within our own galaxy.

The Hubble turning fork diagram

The Hubble "tuning fork" diagram, is also referred to as the Hubble sequence or Hubble classification scheme. This is a graphical representation of different galaxies based on their appearance.

The Large Scale Structure of the Universe

The large-scale structure of the universe is a fascinating field of study that helps us unravel the mysteries of the cosmos and provides valuable clues about the fundamental nature of the universe itself.

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What is the correct order from shortest to longest for these units of measure?Angstrom, astronomical unit, centimeter, kilometer, lightyear, micron, nanometer, parsec?

Answers

The correct order from shortest to longest for these units of measure is as follows:

The correct order from shortest to longest for these units of measure is

1. Nanometer (nm)

2. Angstrom (Å)

3. Micron (μm)

4. Centimeter (cm)

5. Kilometer (km)

6. Astronomical Unit (AU)

7. Lightyear (ly)

8. Parsec (pc)

To give you an idea of the relative magnitudes of these units:

- A nanometer (nm) is equal to 1 billionth of a meter (10^-9 m).

- An Angstrom (Å) is equal to 0.1 nanometers (10^-10 m).

- A micron (μm) is equal to 1 millionth of a meter (10^-6 m).

- A centimeter (cm) is equal to 1 hundredth of a meter (10^-2 m).

- A kilometer (km) is equal to 1,000 meters.

- An astronomical unit (AU) is the average distance between the Earth and the Sun, approximately 150 million kilometers.

- A lightyear (ly) is the distance light travels in one year, approximately 9.46 trillion kilometers.

- A parsec (pc) is a unit of astronomical distance, approximately 3.09 trillion kilometers.

So, the order from shortest to longest represents the increasing magnitude of these units.

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A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 9 ✕ 10−8 C at location <−0.2, 0, 0> m. At location <0, 0.03, 0> m, what is the electric field contributed by the polarization charges on the surface of the metal sphere? (Express your answer in vector form.)

charges = ______________ N/C

Answers

The electric field contributed by the polarization charges on the surface of the metal sphere at location <0, 0.03, 0> m is given by charges =  [tex]{\left(\frac{{9 \times 10^{-8}}}{{4 \pi \varepsilon_0}}\right)} \left(\frac{{\mathbf{r}}}{{|\mathbf{r}|^3}}\right)[/tex] N/C.

To calculate the electric field contributed by the polarization charges on the surface of the metal sphere, we can use the principle of superposition. Each polarization charge can be considered as a point charge contributing to the electric field at the given location.

The electric field created by a point charge q located at position [tex]\mathbf{r'} is given by \mathbf{E} = \frac{1}{{4 \pi \varepsilon_0}} \left(\frac{q}{{|\mathbf{r} - \mathbf{r'}|^3}}\right) (\mathbf{r} - \mathbf{r'})[/tex], where [tex]\varepsilon_0[/tex] is the permittivity of free space.

In this case, we have a neutral solid metal sphere, so the polarization charges only exist on its surface. Let's assume the location of a polarization charge on the surface is given by [tex]\mathbf{r'}[/tex]. The charge q is equal to the product of the surface charge density [tex]\sigma[/tex] and the area element dA at that point, which is [tex]q = \sigma dA[/tex]. The direction of the electric field is given by the vector [tex](\mathbf{r} - \mathbf{r'})[/tex] which points from the location of the polarization charge to the point where we want to calculate the electric field.

By integrating over the entire surface of the sphere, we can determine the total electric field contributed by all the polarization charges. In this case, since the sphere is symmetric, the electric field contributions from all points on the surface cancel out except for the ones along the x-axis. Therefore, we only need to consider the polarization charges located at <−0.2, 0, 0> m. Plugging in the values into the formula, we find that the electric field contributed by the polarization charges on the surface of the metal sphere at location <0, 0.03, 0> m is given by charges =  [tex]{\left(\frac{{9 \times 10^{-8}}}{{4 \pi \varepsilon_0}}\right)} \left(\frac{{\mathbf{r}}}{{|\mathbf{r}|^3}}\right)[/tex] N/C.

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the describes the solar energy emitted by the sun. the arrangement of electromagnetic waves in the spectrum are due to the fact that , also known as sunlight, can simultaneously behave as a and as a wave.

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The electromagnetic spectrum describes the solar energy emitted by the sun.

The arrangement of electromagnetic waves in the spectrum are due to the fact that electromagnetic radiation, also known as sunlight, can simultaneously behave as a particle and as a wave. The electromagnetic spectrum includes different types of electromagnetic radiation, with wavelengths ranging from the shortest gamma rays to the longest radio waves.The sun is a powerful source of energy, and it emits various types of electromagnetic radiation, including visible light, ultraviolet light, and infrared radiation. These different types of radiation have different wavelengths and frequencies, which determine their position on the electromagnetic spectrum. The electromagnetic spectrum is important because it helps scientists understand the behavior of electromagnetic radiation and its interaction with matter.

For example, different types of radiation have different levels of energy, which can cause them to interact differently with materials. In addition, different types of radiation can be used for different applications, such as medical imaging, communication, and energy production.

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1. Suppose you know that the veight of standard poodles is Normally distributed with a standard deviation of 5 pounds. Bob tells you that the average weight of standard poodles is 55 pounds. You take an SRS of 8 poodles and find their average weight to be 52 pounds. (a) Run a two-sided significance test, with a significance level of 5% to see if Bob is correct or not. (b) You have reason to believe that Bob is wrong, and that the average is actually smaller than they ciaim. Run the appropriate onesided significance test with a significance level of 5%. (c) Somebody else believes that Bob is wrong, and that the average is actually larger than the first person claimed. Run the appropriate one-sided significance test with a significance level of 5%. (d) Compare your three answers above.

Answers

the results of the three tests are as follows:

(a) Two-sided test: Fail to reject the null hypothesis. No evidence to suggest a difference in average weight.

(b) One-sided test (smaller): Reject the null hypothesis. Evidence suggests the average weight is smaller.

(c) One-sided test (larger): Fail to reject the null hypothesis. No evidence to suggest the average weight is larger.

These results indicate that there is some evidence to support the claim that the average weight of standard poodles is smaller than 55 pounds,

To perform the significance tests, we can use the t-distribution since the population standard deviation is unknown and the sample size is small (n < 30).

(a) Two-sided significance test:

Null hypothesis (H0): The average weight of standard poodles is 55 pounds.

Alternative hypothesis (Ha): The average weight of standard poodles is not 55 pounds.

Using a significance level of 5% (α = 0.05) and an SRS of 8 poodles with a sample mean of 52 pounds, we can calculate the t-value and compare it to the critical t-value.

The t-value is calculated as (sample mean - population mean) / (sample standard deviation / √n):

t = (52 - 55) / (5 / √8) ≈ -1.897

Degrees of freedom (df) = n - 1 = 8 - 1 = 7

Looking up the critical t-value for a two-sided test with α = 0.05 and df = 7, we find that t_critical ≈ ±2.365.

Since -1.897 falls within the range -2.365 to 2.365, we fail to reject the null hypothesis. There is not enough evidence to conclude that the average weight of standard poodles is different from 55 pounds.

(b) One-sided significance test (smaller):

Null hypothesis (H0): The average weight of standard poodles is 55 pounds.

Alternative hypothesis (Ha): The average weight of standard poodles is less than 55 pounds

Using a significance level of 5% (α = 0.05) and the same sample data, we calculate the t-value and compare it to the critical t-value for a one-sided test.

t = (52 - 55) / (5 / √8) ≈ -1.897

Looking up the critical t-value for a one-sided test with α = 0.05 and df = 7, we find that t_critical ≈ -1.895.

Since -1.897 is slightly less than -1.895, we reject the null hypothesis. There is evidence to suggest that the average weight of standard poodles is smaller than 55 pounds.

(c) One-sided significance test (larger):

Null hypothesis (H0): The average weight of standard poodles is 55 pounds.

Alternative hypothesis (Ha): The average weight of standard poodles is greater than 55 pounds.

Using a significance level of 5% (α = 0.05) and the same sample data, we calculate the t-value and compare it to the critical t-value for a one-sided test.

t = (52 - 55) / (5 / √8) ≈ -1.897

Looking up the critical t-value for a one-sided test with α = 0.05 and df = 7, we find that t_critical ≈ 1.895.

Since -1.897 is less than -1.895, we fail to reject the null hypothesis. There is not enough evidence to conclude that the average weight of standard poodles is greater than 55 pounds.

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when is the emf first one-fourth of its maximum in seconds?

Answers

In order to find out the time when the EMF is first one-fourth of its maximum, we need to know the mathematical expression that defines the EMF of a current-carrying coil. In conclusion, the time when the EMF is first one-fourth of its maximum is t_max(1 - e^(-4)). The mathematical expression for EMF is EMF = -N(dΦ/dt).

According to Faraday's Law of Electromagnetic Induction, EMF can be expressed as:

EMF = -N(dΦ/dt)

where N is the number of turns in the coil, Φ is the magnetic flux passing through the coil and dΦ/dt is the time derivative of magnetic flux.

The EMF of the coil is maximum when the rate of change of magnetic flux is maximum. When the rate of change of magnetic flux is maximum, EMF is also maximum.

Therefore, we can say that:

EMF_max = -N(dΦ/dt)_max

Given that the time when the EMF is first one-fourth of its maximum is t_1/4, we can write:

EMF_1/4 = (1/4)

EMF_max = -N(dΦ/dt)_1/4

We can further simplify the above expression as:

dΦ/dt = -(EMF_1/4)/(N(1/4)(dΦ/dt)_max) = -(4/EMF_max)(EMF_1/4)

Therefore, the time when the EMF is first one-fourth of its maximum is:

t_1/4 = t_max(1 - e^(-4))

where t_max is the time when the EMF is maximum.

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a stunt driver drives a car so fast that it leaves the ground as it tops a hill. if the hill can be approximated by a 125.0- m-radius vertical circle, what speed must the car exceed if it is to leave the ground?

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The car must exceed a speed of 392.93 m/s to leave the ground., when a stunt driver drives a car so fast that it leaves the ground as it tops a hill.

We can calculate the speed needed using the following information: Radius of the hill = 125.0 m. Weight of the car = 1962 kg = 1962 x 9.81 = 19227.42 N Gravitational acceleration, g = 9.81 m/s²

Speed needed by the car to leave the ground can be calculated as follows: Centripetal force provided by the horizontal component of the normal force must be equal to the weight of the car.Centripetal force, Fc = m * v² / rWhere, m = mass of the car, v = speed of the car, r = radius of the hill

Therefore, Fc = 19227.42 Nm * v² / r = 19227.42 Nv² / r = 19227.42 N / (1962 kg * 9.81 m/s²)v² = r * g * (Fc / m)v² = 125.0 m * 9.81 m/s² * (19227.42 N / 1962 kg)v² = 153862.12v = √(153862.12)v = 392.93 m/s

Therefore, the car must exceed a speed of 392.93 m/s to leave the ground.

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