Analyze,how does the inertia of the weights change by increasing their distance to the center - R! Which statement corresponds to Your results? o Inertia is rarely increasing along with R, in less than half of Rmeasurements o Inertia is often increasing along with R, in 50-80% of R measurements o Inertia is mostly, but not always increasing along with R, in at least 80% of R measurements o Inertia is always decreasing with increasing R o Inertia is always increasing along with R

Answers

Answer 1

The inertia of an object is its resistance to change in its state of motion. It is measured by the moment of inertia, which is equal to the mass of the object multiplied by the square of its radius. As the radius of an object increases, its moment of inertia increases, and its inertia decreases.So option 4 is correct.

In the case of the weights, as the distance from the center (R) increases, the moment of inertia increases, and the inertia decreases. This means that it takes less force to accelerate the weights as they move away from the center.

The other options are incorrect.

Option (1) is incorrect because inertia is always increasing with increasing R.

Option (2) is incorrect because inertia is never increasing along with R.

Option (3) is incorrect because inertia is not mostly increasing along with R.

Option (5) is incorrect because inertia is not always increasing along with R.

Therefore  option 4 is correct.

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Related Questions

A rock is dropped off a cliff and falls the first half of thedistance to the ground in t1 seconds. If it fallsthe rest of the distance in t2 seconds, what is thevalue of t2/t1?

Answer: √2 - 1 (1 is not included under the root)

Answers

This is another physics problem that can be solved using the free fall equation, which describes the relationship between the distance traveled (h), time elapsed (t), and acceleration due to gravity (g) for a falling object. The free fall equation in terms of time is:

h = \frac{1}{2}gt^{2}

Let's assume that the total distance from the cliff to the ground is H and the acceleration due to gravity is g. The first half of the distance is H/2 and the second half of the distance is also H/2. The time it takes to fall the first half of the distance is t_{1} and the time it takes to fall the second half of the distance is t_{2}. Using the free fall equation, we can write:

H/2 = \frac{1}{2}gt_{1}^{2}

H/2 = \frac{1}{2}g(t_{1} + t_{2})^{2}

Solving for t_{1} from the first equation, we get:

t_{1} = \sqrt{\frac{H}{g}}

Substituting this into the second equation, we get:

H/2 = \frac{1}{2}g(\sqrt{\frac{H}{g}} + t_{2})^{2}

Expanding and simplifying, we get:

t_{2}^{2} + 2\sqrt{\frac{H}{g}}t_{2} - \frac{H}{g} = 0

Using the quadratic formula, we get:

t_{2} = -\sqrt{\frac{H}{g}} \pm \sqrt{\frac{3H}{g}}

Since t_{2} must be positive, we choose the positive root and get:

t_{2} = -\sqrt{\frac{H}{g}} + \sqrt{\frac{3H}{g}}

Therefore, the ratio of t_{2} to t_{1} is:

\frac{t_{2}}{t_{1}} = \frac{-\sqrt{\frac{H}{g}} + \sqrt{\frac{3H}{g}}}{\sqrt{\frac{H}{g}}} = -1 + \sqrt{3}

This ratio is approximately equal to 0.732.

The ratio of the time taken by a rock to fall the second half of its distance to that taken to fall the first half is given by [tex]$\sqrt{2}-1$[/tex] (1 is not included under the root).

Let's assume that the rock is dropped from a height h. The time taken by the rock to fall the first half of the distance (i.e., h/2) is given by [tex]$t_1=\sqrt{\frac{h}{2g}}$[/tex], where g is the acceleration due to gravity.

Now, let's consider the time taken by the rock to fall the second half of the distance (also h/2). Using the equation of motion, we have [tex]$h/2=\frac{1}{2}gt_2^2$[/tex]. Solving for [tex]$t_2$[/tex], we get [tex]$t_2=\sqrt{\frac{2h}{g}}$[/tex].

Therefore, the ratio of [tex]$t_2$[/tex] to [tex]$t_1$[/tex] is given by:

[tex]\frac{t_2}{t_1}=\frac{\sqrt{\frac{2h}{g}}}{\sqrt{\frac{h}{2g}}}=\sqrt{4}\cdot\frac{\sqrt{\frac{h}{g}}}{\sqrt{h}}=\sqrt{2}.$$[/tex]

Hence, the ratio of the time taken by a rock to fall the second half of its distance to that taken to fall the first half is given by [tex]$\sqrt{2}-1$[/tex] (1 is not included under the root).

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(a) Consider a quasi-static isothermal expansion or compression of an ideal gas, with initial volume V, and final volume V i f (1) What are three thermodynamics coordinates of an ideal gas system? (15

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The thermodynamics coordinates of an ideal gas system include: Pressure (P), Volume (V), and Temperature (T).

In the quasi-static isothermal expansion or compression of an ideal gas, pressure, volume and temperature changes occur. However, the temperature of the gas remains constant since the expansion or compression is isothermal.

Quasi-static process:

A quasi-static process is a process that occurs infinitely slowly such that the gas remains in equilibrium throughout the process. As a result, the system maintains an equilibrium state during the process of expansion or compression.

Ideal gas:

An ideal gas is a hypothetical gas that obeys all the assumptions of the kinetic theory of gases. The kinetic theory of gases suggests that gas molecules do not exert forces on each other and that they are far apart from each other.

The three thermodynamics coordinates of an ideal gas system are Pressure (P), Volume (V), and Temperature (T). These three quantities are called state variables since they define the state of an ideal gas system.

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The Gloria is the second movement of the Mass ____________ as set by Giovanni Pierluigi da Palestrina. The form is _________, the text setting is primarily__ syllabic, and the texture is predominantly ___________.

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The Gloria is the second movement of the Mass Ordinary as set by Giovanni Pierluigi da Palestrina. The form is ABA', the text setting is primarily syllabic, and the texture is predominantly polyphonic.

The Gloria is a musical setting of the Gloria in excelsis Deo, the second section of the Mass Ordinary. The Gloria is a hymn of praise to the Holy Trinity. It is sung on festive occasions such as Christmas, Easter, and major religious holidays.Palestrina's Gloria is written in triple meter and is made up of three sections. The form of the Gloria is ABA', in which the opening section (A) returns after the central section (B). The central section of the Gloria focuses on the story of Christ's birth and is the most musically complex. The third section (A') is a repeat of the first section, which is often accompanied by a festive coda.The text setting of the Gloria is primarily syllabic, meaning that each syllable is assigned a single note, allowing the text to be easily understood. This style is characteristic of Palestrina's music and is intended to enhance the intelligibility of the text.The texture of the Gloria is predominantly polyphonic, meaning that multiple melodies are being sung at the same time. This is a hallmark of Palestrina's style, as he was renowned for his ability to create intricate and beautiful polyphonic music that remained intelligible to the listener.

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a body of mass m is executing simple harmonic motion with an amplitude of 8.0 cm and a maximum acceleration of 100 cm/s2. when the displacement of this body from the equilibrium position is 6.0 cm, the magnitude of the acceleration is approximately

Answers

Answer: Magnitude of acceleration: 937.5 cm/s^2

Explanation:

To find the magnitude of acceleration at a given displacement in simple harmonic motion, we can use the equation:

a = -ω²x

Where:

a is the acceleration,

ω (omega) is the angular frequency, and

x is the displacement from the equilibrium position.

In this case, we are given the amplitude (A) and the maximum acceleration (a_max). The maximum acceleration is equal to ω²A, so we can rearrange the equation to find ω:

ω = √(a_max / A)

Substituting the given values:

a_max = 100 cm/s²

A = 8.0 cm

ω = √(100 cm/s² / 8.0 cm) = √12.5 rad/s

Now we can find the magnitude of acceleration at a displacement of 6.0 cm:

x = 6.0 cm

a = -ω²x = -(12.5 rad/s)² * (6.0 cm) ≈ -937.5 cm/s²

Therefore, the magnitude of the acceleration at a displacement of 6.0 cm is approximately 937.5 cm/s².

DETAILS SERCP11 24.2.P.003. Light at 633 nm from a helium-neon laser shines on a pair of parallel slits separated by 1.31 x 105 m and an interfere HINT (a) Find the angle (in degrees) from the central maximum to the first bright fringe. 0 (b) At what angle (in degrees) from the central maximum does the second dark fringe appear? 0 (c) Find the distance (in m) from the central maximum to the first bright fringe. m Need Help? Read It MY NOTES ASK YOUR TEACHER slits separated by 1.31 x 105 m and an interference pattern is observed on a screen 1.60 m from the plane of the slits. st bright fringe. second dark fringe appear? bright fringe. PRACTICE ANOTHER

Answers

Light at 633 nm from a helium–neon laser shines on a pair of parallel slits separated by 1.57 ✕ 10−5 m and an interference pattern is observed on a screen 2.10 m from the plane of the slits.(1)The angle from the central maximum to the first bright fringe is approximately 2.31°.(2)the angle from the central maximum to the second dark fringe is approximately 3.47°.(3) The distance from the central maximum to the first bright fringe is approximately 0.082 m.

(1)To solve these problems, we can use the equations related to the interference pattern produced by double-slit diffraction.

Given:

Wavelength (λ) = 633 nm = 633 x 10^(-9) mDistance between slits (d) = 1.57 x 10^(-5) mDistance from slits to the screen (L) = 2.10 m

  To find the angle from the central maximum to the first bright fringe, we can use the equation:

sin(θ) = m * λ / d

where:

θ is the angle from the central maximum to the bright fringem is the order of the bright fringe (in this case, m = 1 for the first bright fringe)λ is the wavelength of lightd is the distance between the slits

Plugging in the values:

sin(θ) = 1 * (633 x 10^(-9) m) / (1.57 x 10^(-5) m)

Using a calculator, we find that sin(θ) is approximately 0.0402.

Taking the inverse sine (arc sin) of 0.0402, we find that the angle θ is approximately 2.31°.

Therefore, the angle from the central maximum to the first bright fringe is approximately 2.31°.

(2)  To find the angle from the central maximum to the second dark fringe, we can use the equation:

sin(θ) = (m + 0.5) * λ / d

where:

θ is the angle from the central maximum to the dark fringem is the order of the dark fringe (in this case, m = 1 for the first dark fringe)λ is the wavelength of lightd is the distance between the slits

Plugging in the values:

sin(θ) = (1 + 0.5) * (633 x 10^(-9) m) / (1.57 x 10^(-5) m)

Using a calculator, we find that sin(θ) is approximately 0.0605.

Taking the inverse sine (arcsin) of 0.0605, we find that the angle θ is approximately 3.47°.

Therefore, the angle from the central maximum to the second dark fringe is approximately 3.47°.

(3) To find the distance from the central maximum to the first bright fringe, we can use the equation:

y = L * tan(θ)

where:

y is the distance from the central maximum to the bright fringeL is the distance from the slits to the screenθ is the angle from the central maximum to the bright fringe

Plugging in the values:

y = (2.10 m) * tan(2.31°)

Using a calculator, we find that y is approximately 0.082 m.

Therefore, the distance from the central maximum to the first bright fringe is approximately 0.082 m.

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a cool star is found to have a peak emitted wavelength of 850 nm. what is the stars surface temperature? question 29 options:

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The surface temperature of the cool star is approximately 3412 Kelvin (K) based on its peak emitted wavelength of 850 nm.

The surface temperature of a cool star can be determined based on its peak emitted wavelength. For a star with a peak emitted wavelength of 850 nm, its surface temperature can be calculated using Wien's displacement law.

Wien's displacement law states that the peak wavelength of radiation emitted by a black body is inversely proportional to its temperature. Mathematically, the relationship can be expressed as λ_max = (b / T), where λ_max is the peak wavelength, T is the temperature in Kelvin, and b is Wien's displacement constant.

To find the surface temperature of the cool star, we can rearrange the equation as T = (b / λ_max). The value of Wien's displacement constant is approximately 2.898 × 10⁻³ meters Kelvin (m·K). Converting the given wavelength of 850 nm to meters (0.85 × 10⁻⁶ m), we can substitute these values into the equation to calculate the surface temperature.

T = (2.898 × 10⁻³ m·K) / (0.85 × 10⁻⁶ m) ≈ 3412 Kelvin (K).

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professor brown holds on to the end of the minute hand of a clock atop citv hall. if the minute hand is 4.0 m long. what is the professor's centripetal acceleration?

Answers

The centripetal acceleration of the professor holding onto the end of the minute hand of a clock atop City Hall is 0.00133 m/s².

What is centripetal acceleration?

Centripetal acceleration is the inward force acting on a body moving in a circular path that changes the direction of the velocity of the body and constantly pulls it toward the center of the circle.To determine the professor's centripetal acceleration, we use the formula;

`a= (v²)/r`

Where `a` is the centripetal acceleration, `v` is the velocity, and `r` is the radius. We have the length of the minute hand which is the radius of the circle.

So,`r = 4 m`We need to find the velocity which is given by the formula:

`v= (2πr)/T`

Where `π` is pi (3.14), `r` is the radius, and `T` is the time taken for one complete rotation which is 60 minutes since it's the minute hand.

Therefore;`v = (2 x 3.14 x 4 m) / (60 min x 60 s / 1 min)``v = 0.42 m/s`Substitute `v` and `r` into `a = (v²)/r` to get:`a = (0.42 m/s)² / 4 m``a = 0.00133 m/s²`

Therefore, the centripetal acceleration of the professor holding onto the end of the minute hand of a clock atop City Hall is 0.00133 m/s².

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A balloon weighing 15 grams is filled with helium (density = 0.180 kg/m³) to a volume of 6.0 m³ and attached to a spring of force constant 120 N/m. It Determine the extension of the spring, L, when the balloon is in equilibrium. The density of air is 1.29 kg/m³.

Answers

The extension of the spring, L, when the balloon is in equilibrium is 0.63 m.

Weight of the balloon = 15 g = 0.015 kg

Density of helium = 0.180 kg/m³

Volume of balloon = 6.0 m³

Force constant of spring, k = 120 N/m

Density of air = 1.29 kg/m³

Extension of the spring, L, when the balloon is in equilibrium using Hooke's law, F = kx

Let's first find the buoyancy force on the balloon when it is filled with helium and determine its weight. Buoyancy force = weight of the air displaced by the balloon

Buoyancy force = Density of air × volume of the balloon × gravitational acceleration = 1.29 kg/m³ × 6.0 m³ × 9.8 m/s² = 75.768 N

Weight of the balloon = Mass of the balloon × gravitational acceleration= 0.015 kg × 9.8 m/s² = 0.147 N

Therefore, the net force acting on the balloon when it is filled with helium is given by

Net force = Buoyancy force - Weight of balloon = 75.768 N - 0.147 N = 75.621 N

This net force acts upward on the balloon.

Now, using Hooke's law, we can determine the extension of the spring, L, when the balloon is in equilibrium.

F = kx, where F is the net force acting on the balloon, and k is the force constant of the spring.

Substituting the values of F and k, we get75.621 N = 120 N/m × L

Therefore,

L = 75.621 N / 120 N/m = 0.63 m

Therefore, the extension of the spring, L, when the balloon is in equilibrium is 0.63 m.

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The force F between two parallel wires carrying electric currents is inversely proportional to the distance d between the wires. If a force of 0.750 N exists between wires that are 1.75 cm apart, what is the force between them if they are separated by 2.50 cm?

Answers

the force between the two wires if they are separated by 2.50 cm is 0.525 N.

Given that force F between two parallel wires carrying electric currents is inversely proportional to the distance d between the wires and that a force of 0.750 N exists between wires that are 1.75 cm apart and that we are supposed to find the force between them if they are separated by 2.50 cm.

Let the initial force be F₁ and the initial distance be d₁.

Therefore, we can write the relationship between force and distance as;

F₁d₁ = F₂d₂

Where

;F₁ = 0.750 N (initial force)

d₁ = 1.75 cm (initial distance)

F₂ = ? (force at new distance)

d₂ = 2.50 cm (new distance)

Let us find F₂;F₁d₁ = F₂d₂F₂ = F₁d₁/d₂

Now substitute the values we know;

F₂ = (0.750 N x 1.75 cm) / 2.50 cmF₂ = 0.525 N

Therefore, the force between the two wires if they are separated by 2.50 cm is 0.525 N.

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An alpha particles is shot with a speed of 2*10^7 m/s
directly toward the nucleus of gold atom. What is the distance of
closest approach to the nucleus?

Answers

An alpha particles is shot with a speed of 2*10^7 m/s directly toward the nucleus of gold atom.  The distance of closest approach to the nucleus is approximately 1.27 × 10^-14 meters.

To calculate the distance of closest approach to the nucleus, we can use the concept of the Rutherford scattering formula. The Rutherford scattering formula is given by:

R = (k * Z1 * Z2 * e^2) / (2 * π * ε₀ * m * v₀²)

Where:

R is the distance of closest approach

k is Coulomb's constant (9 × 10^9 N m²/C²)

Z1 and Z2 are the atomic numbers of the particles involved

e is the elementary charge (1.6 × 10^-19 C)

ε₀ is the permittivity of free space (8.85 × 10^-12 F/m)

m is the mass of the alpha particle

v₀ is the initial velocity of the alpha particle

Given:

Z1 = 2 (atomic number of alpha particle)

Z2 = 79 (atomic number of gold atom)

v₀ = 2 × 10^7 m/s

Substituting the values into the formula:

R = (9 × 10^9 N m²/C² * 2 * 79 * (1.6 × 10^-19 C)^2) / (2 * π * 8.85 × 10^-12 F/m * (6.64 × 10^-27 kg) * (2 × 10^7 m/s)^2)

Calculating the value:

R = (9 × 10^9 N m²/C² * 2 * 79 * (2.56 × 10^-38 C²)) / (2 * π * 8.85 × 10^-12 F/m * (6.64 × 10^-27 kg) * 4 × 10^14 m²/s²)

R = (9 × 10^9 * 2 * 79 * 2.56 × 10^-38) / (2 * π * 8.85 × 10^-12 * 6.64 × 10^-27 * 4 × 10^14)

R ≈ 1.27 × 10^-14 meters

Therefore, the distance of closest approach to the nucleus is approximately 1.27 × 10^-14 meters.

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18. How many electrons pass through a 20-22 resistor in 10 minutes if there is a potential difference of 32 V across its ends? A. 3.2 × 1021 B. 6.0 × 1021 C. 1.6 × 1021 D. 2.0 × 1021 E. 9.0 × 102

Answers

The number of electrons passing through the resistor is approximately 3.07 × 10^21. So option A  is correct.

To calculate the number of electrons passing through a resistor, we need to determine the charge passing through the resistor and then convert it to the number of electrons.

The charge passing through the resistor (Q) can be calculated using the formula:

Q = V ×t

Where Q is the charge, V is the potential difference or voltage  across the resistor, and t is the time.

Given:

Potential difference (V) = 32 V

Time (t) = 10 minutes = 10 * 60 seconds = 600 seconds

Q = 32 V × 600 s

Q = 19,200 Coulombs

To convert the charge (Q) to the number of electrons, we need to divide it by the elementary charge (e).

1 Coulomb (C) = 6.242 × 10^18 elementary charges (e)

Number of electrons (n) = Q / e

n = 19,200 C / (6.242 × 10^18 e)

n ≈ 3.07 × 10^21 electrons

Therefore, the number of electrons passing through the resistor is approximately 3.07 × 10^21.Therefore option  A is correct.

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4]. Water in an electric kettle connected
to 220V supply takes 5minutes to reach
its boiling point. How long would it have
taken if the supply voltage had fallen to
200V?

Answers

If the supply voltage had fallen to 200V, it would take 5.5 minutes for water in an electric kettle to reach its boiling point.

Let the time taken when the supply voltage is 220V be t₁. Therefore, t₁ = 5 mins

The power of the kettle is given by P₁ = V₁I where V₁ is the voltage, I is the current.

Potential difference V is inversely proportional to the current I.

P₁ = V₁I => I = P₁/V₁

Therefore, when the voltage is reduced to 200V, the current will beI = P₁/V₂ where V₂ is the new voltage.

Now let the time taken for water to reach boiling point when voltage is 200V be t₂.

Then V₁I₁t₁ = V₂I₂t₂

But we have I₁ = I₂

So, V₁t₁ = V₂t₂t₂

= (V₁/ V₂) * t₁t2

= (220/200) * 5t₂

= 5.5 mins

Therefore, if the supply voltage had fallen to 200V, it would take 5.5 minutes for water in an electric kettle to reach its boiling point.

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Mobile carriers use the n78 band at 3.5×109[Hz]. What is the
wavelength at this frequency? Show solution.
A. 1.16×10−1[m]
B. 1.16×101[m]
C. 8.57×102[m]
D. 8.57×10−2[m]

Answers

The wavelength of mobile carriers that use the n78 band at 3.5×10^9[Hz] is 8.57×10−2[m].

The wavelength of a signal is defined as the distance between two corresponding points of the same phase on a given wave.

It can be determined using the formula:λ=cswhere λ is the wavelength, c is the speed of light, and s is the frequency. Mobile carriers use the n78 band at 3.5×10^9[Hz], which means the frequency of the signal is given as s=3.5×10^9[Hz]. The speed of light is approximately 3×10^8[m/s].

Hence, substituting these values into the above formula, we get:λ=3×10^8/3.5×10^9=8.57×10−2[m].

Therefore, the wavelength of the mobile carriers that use the n78 band at 3.5×10^9[Hz] is 8.57×10−2[m].

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1. 30 in 2. 120 Km/h 3. 60 mi/h 4. 53785 g 5. 358235 ms 6. 95 ft 7. 0.95786 Mm 8. 5ft 6in 9. 974 cm/min 10. 863289 μm Conversion Factors 1 m = 39.37 in 1 mi= 1609 m 1m = 3.28 ft HW 6 Unit Conversions

Answers

Unit conversions are an important part of solving problems in physics and mathematics. It is necessary to learn different conversion factors to be able to convert one unit to another unit.

 (1)  30 inches to meters:

   Conversion factor: 1 inch = 0.0254 meters

   Calculation: 30 inches * 0.0254 meters/inch = 0.762 meters

(2)  120 km/h to m/s:

   Conversion factor: 1 km/h = 0.2778 m/s

   Calculation: 120 km/h * 0.2778 m/s = 33.336 m/s

(3) 60 mi/h to km/h:

   Conversion factor: 1 mi/h = 1.609 km/h

   Calculation: 60 mi/h * 1.609 km/h = 96.54 km/h

(4) 53785 grams to kilograms:

   Conversion factor: 1 gram = 0.001 kilogram

   Calculation: 53785 grams * 0.001 kilogram/gram = 53.785 kilograms

(5)  358235 milliseconds to seconds:

   Conversion factor: 1 millisecond = 0.001 second

   Calculation: 358235 milliseconds * 0.001 second/millisecond = 358.235 seconds

(6)  95 feet to meters:

   Conversion factor: 1 foot = 0.3048 meters

   Calculation: 95 feet * 0.3048 meters/foot = 28.956 meters

(7)   0.95786 megameters to kilometers:

   Conversion factor: 1 megameter = 1000000 kilometers

   Calculation: 0.95786 megameters * 1000000 kilometers/megameter = 957860 kilometers

 (8)  5 feet 6 inches to centimeters:

   Conversion factor: 1 foot = 30.48 centimeters, 1 inch = 2.54 centimeters

   Calculation: 5 feet * 30.48 centimeters/foot + 6 inches * 2.54 centimeters/inch = 167.64 centimeters

 (9)  974 centimeters per minute to meters per second:

   Conversion factor: 1 minute = 60 seconds, 1 centimeter = 0.01 meter

   Calculation: 974 centimeters/minute * 0.01 meter/centimeter / 60 seconds/minute = 0.1623 meters/second

(10) 863289 micrometers to millimeters:

   Conversion factor: 1 micrometer = 0.001 millimeter

   Calculation: 863289 micrometers * 0.001 millimeter/micrometer = 863.289 millimeters

These are the conversions for the given values using the provided conversion factors.

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Homework 1 1 5 points Match the following quantities with their SI base units. For any derived quantity, both the derived unit and its associated base units are given. N Force m Length Mass S Time m/s

Answers

The quantities and their corresponding SI base units are as follows:

Force (N): Newton , Length (m): meter ,Mass (kg): kilogram ,Time (s): second ,Speed (m/s): meter per second

In the International System of Units (SI), there are seven base units that represent the fundamental quantities in physics. These base units form the foundation for all other units and measurements.

Force (N): The base unit for force is the Newton (N). It is derived from the base units of mass (kg), length (m), and time (s) and is defined as the amount of force required to accelerate a mass of 1 kilogram at a rate of 1 meter per second squared.

Length (m): The base unit for length is the meter (m). It is the fundamental unit for measuring distance and is defined as the distance traveled by light in a vacuum during a specific time interval.

Mass (kg): The base unit for mass is the kilogram (kg). It is defined as the mass of the International Prototype of the Kilogram, a platinum-iridium cylinder kept at the International Bureau of Weights and Measures.

Time (s): The base unit for time is the second (s). It is defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between two hyperfine levels of the cesium-133 atom.

Speed (m/s): Speed is a derived quantity, and its base units are meters per second (m/s). It represents the rate at which an object covers a distance in a given amount of time.

The quantities and their associated SI base units are:

Force (N) - Newton

Length (m) - meter

Mass (kg) - kilogram

Time (s) - second

Speed (m/s) - meter per second

These base units form the basis for measuring the respective quantities and are widely used in scientific and everyday applications.

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40. What is the polarization angle if the unpolarized sun light is incident to a smooth glass (n=1.55) surface? a. 57.2 degrees c. 75.2 degrees d. 47.2 degrees b. 67.3 degrees

Answers

The polarization angle, which is the angle between the reflected polarized light and the plane of incidence, is given by θ_p ≈ 55.22 degrees.So option a is correct.

The polarization angle can be determined using Brewster's law, which states that when light is incident on a medium at a specific angle, known as the Brewster's angle (θ_B), the reflected light becomes completely polarized perpendicular to the plane of incidence.

Brewster's law can be expressed as:

tan(θ_B) = n2/n1

where n2 is the refractive index of the second medium (in this case, air with n2 = 1.00) and n1 is the refractive index of the first medium (glass with n1 = 1.55).

Let's calculate the Brewster's angle (θ_B):

tan(θ_B) = 1.00/1.55

θ_B = arc tan(1.00/1.55)

Using a calculator, we find:

θ_B ≈ 34.78 degrees

The polarization angle, which is the angle between the reflected polarized light and the plane of incidence, is given by:

θ_p = 90 - θ_B

θ_p = 90 - 34.78

θ_p ≈ 55.22 degrees

Therefore option a is correct.

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Page Introduction: Boyle's Law describes the relationship between a volume of gas and its pressure when held at constant temperature. The pressure is inversely proportional to the volume. This means a

Answers

Given amount of gas will have a higher pressure if its volume is decreased, and vice versa, as long as the temperature remains constant.

Boyle's Law, named after the physicist Robert Boyle, states that the pressure of a gas is inversely proportional to its volume at a constant temperature. In other words, when the volume of a gas decreases, its pressure increases, and when the volume increases, the pressure decreases, as long as the temperature remains constant.

This behavior can be understood by considering the movement of gas molecules. When the volume of a gas is reduced, the same number of molecules are confined to a smaller space, leading to more frequent collisions with the container walls. These collisions exert a greater force per unit area, resulting in an increase in pressure. Conversely, when the volume is increased, the gas molecules have more space to move around, reducing the frequency of collisions and thus lowering the pressure.

Mathematically, Boyle's Law can be expressed as P₁V₁ = P₂V₂, where P₁ and V₁ represent the initial pressure and volume, and P₂ and V₂ represent the final pressure and volume. This equation shows the inverse relationship between pressure and volume.

In summary, Boyle's Law states that the pressure of a gas is inversely proportional to its volume at a constant temperature. Decreasing the volume of a gas will cause an increase in pressure, while increasing the volume will result in a decrease in pressure, as long as the temperature remains constant.

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a light-year is considered to be the distance that light travels in 1 year. the speed of light is about 300,000 kilometers per second. if 1 earth year is 31,536,000 seconds, how far does light travel in 1 earth year? represent your answer using scientific notation.

Answers

A light-year is considered to be the distance that light travels in one year. The speed of light is about 300,000 kilometers per second.

If 1 Earth year is 31,536,000 seconds, how far does light travel in 1 Earth year?

The total distance that light travels in 1 Earth year can be found by multiplying the speed of light by the number of seconds in 1 year, which is represented by the following formula:

Distance = speed x time

Distance = 300,000 km/s x 31,536,000 s= 9.46 × 10¹² km (in scientific notation)

Thus, the distance light travels in 1 Earth year is 9.46 × 10¹² kilometers.

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_______ is the point in the object around which its weight is evenly distributed.

Answers

Answer:

CENTER OF GRAVITY is the point in the object around which its weight is evenly distributed.

Explanation:

An object's center of gravity (CG) is the equilibrium point where its constituent parts are uniformly distributed. In this situation, the object may behave as though its entire weight were concentrated at the center of gravity (CG).

Applications include the concept that a weighted object always rotates freely about its center of mass and that a weighted object will fall over if its center of gravity is beyond its base of support. Additionally, the center of gravity is where the most force is applied.

The point in the object around which its weight is evenly distributed is known as the center of gravity.

It is also referred to as the center of mass. The center of gravity is the point around which the mass of an object is evenly distributed in all directions. There are different ways to find the center of gravity of an object. However, one common method involves suspending the object from different points and then marking the vertical line. The intersection of these lines is the center of gravity. The center of gravity has applications in physics and engineering. For instance, in the design and construction of buildings, it is essential to determine the center of gravity to ensure the stability and safety of the structure. In summary, the center of gravity is an important concept in physics and engineering that helps in understanding the distribution of weight and stability of objects.

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using conservation of energy and momentum equations show that a free electron cannot absorb a photon.

Answers

To show that a free electron cannot absorb a photon, we can examine the conservation of energy and momentum during the interaction.

To show that a free electron cannot absorb a photon, we can examine the conservation of energy and momentum during the interaction.

The conservation of energy states that the total energy before and after the interaction must remain constant. The energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency of the photon. The energy of a free electron is given by E = p² / (2m), where p is the momentum of the electron and m is its mass.

Let's assume that the free electron absorbs a photon. In this case, the energy of the electron before absorption is E_initial = p_initial² / (2m), and after absorption, it becomes E_final = p_final² / (2m) + hf.

Now let's consider the conservation of momentum. The momentum of a photon is given by p = hf / c, where c is the speed of light. The initial momentum of the electron is p_initial, and after absorption, it becomes p_final.

Applying conservation of energy:

E_initial = E_final

p_initial² / (2m) = p_final² / (2m) + hf

Applying conservation of momentum:

p_initial = p_final + p_photon

p_initial = p_final + hf / c

Substituting the expression for p_initial in terms of p_final and hf:

p_final + hf / c = p_final / (2m) + hf

Simplifying the equation:

2mhf + 2mcp_final = [tex]c^{2p-final^{2}[/tex]  + 2mhf  

Canceling out common terms:

2mcp_final = [tex]c^{2p-final^{2}[/tex]

Simplifying further and Dividing by c:

2m = c p_final

Now, if we examine the equation, we see that the left side (2m) is a constant determined by the mass of the electron, while the right side (c p_final) depends on the momentum of the electron. Since the left side is a constant and the right side depends on p_final, there is no possible value of p_final that satisfies this equation. Therefore, we can conclude that a free electron cannot absorb a photon while conserving energy and momentum.

This result is consistent with the principles of quantum mechanics, where the absorption or emission of a photon by an electron is governed by quantum transitions between energy levels, such as those occurring in atoms or solid-state systems.

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If you observe a light particle at rest floating in the air and you know that the particle is electrically charged, we can say that the force that lifts the particle (compensates for gravity):

a. can't be magnetic
b. can be magnetic
c. it is magnetic
d. None of the above

Answers

If you observe a light particle at rest floating in the air and you know that the particle is electrically charged, we can say that the force that lifts the particle can e magnetic. Option (B) is correct.

When observing a light particle at rest floating in the air, there are different types of forces that act on it. In the case where the particle is electrically charged, the particle will experience an electrical force as well as a gravitational force.

The electrical force is produced as a result of the interaction between the electrically charged particle and other charged particles in its surrounding environment. On the other hand, the gravitational force is the force exerted by the earth's gravitational field on the particle.

However, in order for the particle to be lifted and remain suspended in the air, there needs to be a force that counteracts the force of gravity. This force is known as the "upward force" and is provided by the air resistance acting on the particle.

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a 35.7 kg girl and a 57.6 kg boy are on the surface of a frozen lake, 11.5 m apart. using a rope, the girl exerts a horizontal 4.35 n force on the boy, pulling him toward her. calculate the magnitude of the girl's acceleration.

Answers

The magnitude of the girl's acceleration is determined as 0.12 m/s².

What is the magnitude of the girl's acceleration?

The magnitude of the girl's acceleration is calculated by applying Newton's second law of motion as follows;

F (net) = ma

where;

m is the mass of the girla is the acceleration of the girl

The mass of the girl = 35.7 kg

The net force on the girl = 4.35 N

The magnitude of the girl's acceleration is calculated as;

a = F / m

a = 4.35 N / 35.7 kg

a = 0.12 m/s²

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A swimmer swims for 68 m on the bearing 036°. beast of her starting point? How far is the swimmer a north

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The swimmer is approximately 39.22 meters north of her starting point. This can be determined by finding the north component of the displacement using trigonometry. The north component is calculated by multiplying the distance swum by the sine of the bearing.

To determine how far the swimmer is from her starting point in a north direction, we need to find the north component of the displacement.

Given:

Distance swum (d) = 68 m

Bearing (θ) = 036°

To find the north component, we can use trigonometry.

North Component = d * sin(θ)

North Component = 68 m * sin(36°)

North Component ≈ 39.22 m

Therefore, the swimmer is approximately 39.22 m north of her starting point.

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2-

Which of these is Not a type of electromagnetic radiation?

electrical current from a 9 volt battery

visible yellow light

x-rays

3-

Which of the following lists is correctly ordered from shortest to longest wavelength?

radio, infrared (IR), ultraviolet (UV), gamma rays

gamma rays, UV, IR, radio waves

gamma rays, UV, radio waves, IR

4-

The bright red emission line for hydrogen ( H-alpha line), results from the drop (transition) of its electron from the n = 3 to n = 2 level.

True

False

Answers

Electrical current from a 9 volt battery is not a type of electromagnetic radiation.

The following list is correctly ordered from shortest to longest wavelength:

gamma rays, UV, radio waves, IR.

The statement "The bright red emission line for hydrogen ( H-alpha line), results from the drop (transition) of its electron from the n = 3 to n = 2 level" is True. Electromagnetic radiation consists of oscillating electric and magnetic fields that travel through space at the speed of light. It includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Electrical current from a 9 volt battery is not a type of electromagnetic radiation. It is a flow of electric charge, which is not an oscillating electric and magnetic field.

The following list is correctly ordered from shortest to longest wavelength: gamma rays, UV, radio waves, IR. Gamma rays have the shortest wavelength, followed by UV, radio waves, and then IR.

The statement "The bright red emission line for hydrogen ( H-alpha line), results from the drop (transition) of its electron from the n = 3 to n = 2 level" is True. When the electron of a hydrogen atom drops from a higher energy level to a lower energy level, it emits a photon of light.

The energy of the photon depends on the difference in energy between the two levels. The H-alpha line is a specific emission line that results from the transition of an electron from the n = 3 to n = 2 energy level.

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if the ectopic impulse arises from the middle of the right atrium the p' wave is:

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An ectopic heartbeat is an irregular heartbeat that happens when the heart's sinoatrial node (SA node), which is its normal heartbeat pacemaker, is disrupted. Ectopic beats originate from a location outside of the SA node, disrupting the normal heart rhythm. When it comes to the various types of ectopic beats, the most common is premature ventricular contraction (PVC).

If the ectopic impulse arises from the middle of the right atrium, the P wave will be abnormal. This occurs when the heart's ventricles experience an unexpected electrical impulse, causing them to contract prematurely. The P wave is a wave that appears on an electrocardiogram (ECG) and represents the electrical activity of the atria. The sinoatrial node generates a normal P wave, which spreads through both atria and then travels to the atrioventricular node, which slows the impulse and transmits it to the ventricles. P’ waveIf the ectopic impulse arises from the middle of the right atrium, the P' wave is abnormal.

As a result, the ECG can display the following:P waves with a single, smooth contour that are narrower than normal, P waves that have a pointed apex and are taller than normal, and P waves that merge with other waves, making them indistinguishable on the ECG.P prime waves, which are visible on an ECG, are related to supraventricular beats. They're usually seen in the early part of a supraventricular tachycardia event, which is a fast heart rate originating from the atria.

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A 60-ton locomotive is to be used to haul coal from a loaded sidetrack 6400 ft to the outside. Trips are to be designed to accelerate at a rate of 0.15 mphps up a 2% grade. If 12-ton capacity cars with a tare weight of 6 tons are used, how many locomotives will be required to haul a shift tonnage of 10000 if the locomotives average 15 mph with 4-min turnaround time at each end? The tread is made from cast iron and plain bearings are employed throughout. The operating time per shift is approximately 7 hr (1 ton = 2000 pounds).

Answers

12000 pounds locomotives need to accelerate at a rate of 0.15 mphps up a 2% grade, averaging 15 mph with a 4-minute turnaround time at each end. The operating time per shift is approximately 7 hours.

In order to calculate the number of locomotives required, we need to consider several factors. First, we calculate the net weight of each loaded car by subtracting the tare weight from the capacity weight: 12 tons - 6 tons = 6 tons. Next, we convert the net weight to pounds: 6 tons * 2000 pounds/ton = 12,000 pounds.

To determine the force required to accelerate the train up the 2% grade, we multiply the net weight of each car by the acceleration rate: 12,000 pounds * 0.15 mphps = 1800 mhp (mhp = thousand pounds).

Considering the locomotives' average speed of 15 mph and the 4-minute turnaround time at each end, the effective operating time per hour is 60 minutes / (15 mph + (4 min/60 min)) = 3.75 minutes/mile.

Given that the total distance to be covered is 6400 ft, which is approximately 1.2121 miles (6400 ft / 5280 ft/mile), the total operating time required is 1.2121 miles * 3.75 minutes/mile = 4.5455 minutes.

Since the operating time per shift is approximately 7 hours or 420 minutes, we can calculate the number of trips needed: 420 minutes / (4.5455 minutes/trip) = 92.4 trips.

Since each trip requires a locomotive, we round up to the nearest whole number. Therefore, a minimum of 93 locomotives will be required to haul a shift tonnage of 10,000.

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as two solutions are mixed in a beaker the beaker gets warm and a white solid fall

Answers

The warm beaker and the white solid that falls are both evidence of a chemical reaction.

What is chemical reaction?

A chemical reaction entails the progression wherein one or more substances, known as reactants, undergo a conversion into one or more distinct substances, referred to as products.

The reactants and products are intricately interconnected, as they are bound together by chemical forces. Within the reactants, the chemical bonds are disassembled, while novel chemical bonds are fashioned within the products. This phenomenon is identified as a chemical metamorphosis, signifying a transformative endeavor.

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Complete question:

As two solutions are mixed in a beaker the beaker gets warm and a white solid fall. what does this indicate?

A 5.0-m-long ladder has mass 13.5 kg and is leaning against a frictionless wall, making a 66° angle with the horizontal. Review | Constants Part A If the coefficient of friction between the ladder and ground is 0.42, what is the mass of the heaviest person who can safely ascend to the top of the ladder? (The center of mass of the ladder is at its center.) Express your answer using two significant figures. 15. ΑΣΦ ? mmaz Submit Request Answer kg

Answers

The

mass

of the heaviest person who can safely ascend to the top of the ladder is 13.5 kg.

To solve this problem, we need to analyze the

forces

acting on the ladder and find the maximum mass of a person that can safely ascend to the top.

Let's consider the forces acting on the ladder:

Weight: The ladder has a mass of 13.5 kg, so its weight can be calculated as W_ ladder = m_ ladder * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Normal force: The ladder is in contact with the ground, so there is a normal force acting perpendicular to the ground.

Frictional force: The coefficient of

friction

between the ladder and the ground is given as 0.42. The frictional force can be calculated as F_ friction = coefficient of friction * normal force.

Horizontal component of the force due to the weight: The weight of the ladder can be resolved into two components - a vertical component and a horizontal component. The horizontal component of the weight will push the ladder away from the wall.

Force exerted by the wall: The wall exerts a force on the ladder perpendicular to its surface, preventing it from sliding down.

For the ladder to be in equilibrium, the sum of the forces in the horizontal direction and the sum of the forces in the vertical direction should both be zero.

Let's calculate the forces:

Horizontal forces:

Force exerted by the wall = 0 (frictionless wall)

Vertical forces:

Normal force - weight of the ladder = 0

Normal force = W_ ladder

Now, let's calculate the maximum mass of the person who can safely ascend to the top. We'll consider the point where the person is at the top of the ladder as the center of mass.

The person exerts a downward force due to their

weight,

and this force should be balanced by the upward normal force provided by the ladder. The maximum mass of the person can be calculated as:

Maximum mass of the person = Normal force / g

Substituting the value of the normal force, we have:

Maximum mass of the person = W_ ladder / g

Plugging in the given values, we get:

Maximum mass of the person = (13.5 kg * 9.8 m/s^2) / 9.8 m/s^2 = 13.5 kg

Therefore, the mass of the heaviest person who can safely ascend to the top of the

ladder

is 13.5 kg.

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A pulse can be described as a single wave disturbance that moves through a medium. Consider a pulse that is defined at time t=0.00 st=0.00 s by the equation y(x)=12x2+2 my(x)=12x2+2 m centered around x=0 mx=0 m. The pulse moves with a velocity of v=2.8 m/sv=2.8 m/s in the positive xx-direction.

What is the amplitude of the pulse?
A=A= m
Where is the pulse centered at time t=5.00 st=5.00 s?
x=x= m

A pulse can be described as a single wave disturbance that moves through a medium. Consider a pulse that 1 is defined at time t = 0.00 s by the equation y(x) =1/(2x^2+2)m centered around x = 0 m. The pulse 2.42 + 2 moves with a velocity of v = 2.8 m/s in the positive x-direction.
a. What is the amplitude of the pulse? A=____ m
b. Where is the pulse centered at time t = 5.00 s? m = ____

Answers

The amplitude of the pulse is A =3.464 m, and the pulse is centered at x = 14.00m at time t = 5.00 s.

Explanation:

To determine the amplitude of the pulse, we can look at the given equation for y(x). The amplitude represents the maximum displacement from the equilibrium position. The amplitude of the pulse, we look at the equation y(x) = 12x² + 2 m. The amplitude is the maximum displacement from the equilibrium position. In this case, the coefficient of x² is 12, so the amplitude is the √12, which is approximately 3.464 m.

To determine the center of the pulse at time t = 5.00 s, we need to consider the velocity of the pulse. The pulse moves with a velocity of v = 2.8 m/s in the positive x-direction. Since the pulse is centered around x = 0 m at t = 0.00 s, we can use the formula x = vt to find the center position at a given time. Plugging in the values, we have x = (2.8 m/s)(5.00 s) = 14.00 m. Therefore, the pulse is centered at x = 14.00 m at time t = 5.00 s.

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Which of the following statements comparing halo stars to our Sun is not true?

a. Most stars in the halo have cooler surface temperatures than the Sun.
b. Most stars in the halo are less luminous than the Sun.
c. Most stars in the halo contain a much lower percentage of heavy elements than the Sun.
d. Most stars in the halo have either died or are in their final stages of life, while the Sun is only in about the middle of its lifetime.

Answers

Halo stars are called metal-poor stars, and they are part of the halo of the Milky Way. These stars are much older than the stars that we see in the Milky Way's disk. The disk is a thin layer of stars that includes the sun. The halo stars are older, which means that they have a low metal content.

The following statement that compares halo stars to our sun is not true "Most stars in the halo have either died or are in their final stages of life, while the Sun is only in about the middle of its lifetime. Their low metal content implies that they have few elements that are heavier than helium. These stars are also cooler and less luminous than the sun. The percentage of heavy elements in most halo stars is much lower than that in the sun, as the third option claims. The last option that indicates that most halo stars have either died or are in their final stages of life, while the sun is only in the middle of its lifetime, is untrue because most halo stars are still alive and shining. Hence, the correct answer is option d.

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