the force applied to a 0.4m by 0.8m break pad produces a pressure of 500 N/m².Calculate the force applied to the break pad.​

Answers

Answer 1

The force applied to the brake pad is 160 Newtons.

How to solve for the force

To calculate the force applied to the brake pad, we need to multiply the pressure by the area.

Given:

Pressure = 500 N/m²

Area = 0.4 m * 0.8 m = 0.32 m²

The formula to calculate force is:

Force = Pressure * Area

Substituting the given values:

Force = 500 N/m² * 0.32 m²

Force = 160 N

Therefore, the force applied to the brake pad is 160 Newtons.

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Related Questions

a) The phase velocity of surface waves of wavelength '2' on a liquid of density 'p' and SлT gλ surface tension "T' is given by v= + Deduce the expression for the group 2p 8T velocity in terms of phase velocity? b) An electron is accelerated through a potential of 50 kV. Calculate the wavelength of (10 Marks) the matter wave associated with the electron?

Answers

a) The expression for the group velocity of surface waves on a liquid with wavelength λ, density ρ, and surface tension T can be deduced from the phase velocity formula v = √(gλ/2π) as follows: v_group = v_phase / 2π

b) To calculate the wavelength of the matter wave associated with an electron accelerated through a potential of 50 kV, we can use the de Broglie wavelength formula: λ = h / √(2meV), where h is the Planck's constant, me is the mass of the electron, and V is the potential difference.

a) The phase velocity of surface waves on a liquid with wavelength λ, density ρ, and surface tension T is given by the formula

v_phase = √(gλ/2π), where g is the acceleration due to gravity.

To find the group velocity, we divide the phase velocity by 2π, resulting in the expression:

v_group = v_phase / 2π.

b) According to the de Broglie wavelength formula, the wavelength (λ) of a matter wave associated with a particle can be calculated using λ = h / √(2meV),

where h is the Planck's constant (approximately 6.626 x 10^-34 J·s), me is the mass of the electron (approximately 9.109 x 10^-31 kg),

and V is the potential difference (50 kV = 50,000 volts = 50,000 J/C).

Plugging in the values, we have λ = (6.626 x 10^-34 J·s) / √(2(9.109 x 10^-31 kg)(50,000 J/C)).

Simplifying the expression gives λ ≈ 1.227 x 10^-10 meters.

Therefore, the wavelength of the matter wave associated with the electron is approximately 1.227 x 10^-10 meters.

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A car engine receives 210 kW from a heat source to deliver 55 kW of power to the wheels while rejecting heat to the surroundings at 20oC. It is known that the maximum thermal efficiency this engine can achieve is 40 percent. Determine (a) the thermal efficiency of the engine, (b) the maximum power that can be produced by the engine, and (c) the temperature of the heat source.

Answers

The thermal efficiency of the engine is 26.19%.  the temperature of the heat source is approximately 33.33°C.

(a) The thermal efficiency of an engine is given by the ratio of the useful work output to the heat input. In this case, the useful work output is 55 kW, and the heat input is 210 kW. Therefore, the thermal efficiency can be calculated as (useful work output / heat input) * 100%.

Thermal efficiency = (55 kW / 210 kW) * 100% = 26.19%

(b) The maximum power that can be produced by the engine is when the thermal efficiency is at its maximum. We are given that the maximum thermal efficiency is 40 percent. Therefore, we can use this maximum efficiency value to calculate the maximum power output.

Maximum power output = (Maximum thermal efficiency * Heat input) = (40% * 210 kW) = 84 kW

(c) To determine the temperature of the heat source, we can use the Carnot efficiency formula, which relates the temperatures of the hot and cold reservoirs to the thermal efficiency. The Carnot efficiency is given by the formula:

Carnot efficiency = 1 - (Tc / Th)

Where Tc is the temperature of the cold reservoir (20°C) and Th is the temperature of the hot reservoir (heat source).

Rearranging the formula, we have:

Th = Tc / (1 - Carnot efficiency) = (20°C) / (1 - 0.40) = 33.33°C

Therefore, the temperature of the heat source is approximately 33.33°C.

In summary, the thermal efficiency of the engine is 26.19%, the maximum power output is 84 kW, and the temperature of the heat source is approximately 33.33°C.

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at what displacement from equilibrium does the ball have half its maximum velocity?

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At a displacement equal to half the amplitude (D = A/2), the ball will have half its maximum velocity.

The displacement from equilibrium at which a ball has half its maximum velocity depends on the specific system and its characteristics. However, in a simple harmonic motion system (e.g., a mass-spring system), the displacement from equilibrium at which the ball has half its maximum velocity is equal to half the amplitude of the motion.

In simple harmonic motion, the velocity of the ball is maximum at the equilibrium position (zero displacement) and decreases as the ball moves away from the equilibrium position. The velocity is zero at the maximum displacement (amplitude) and then reverses direction.

Therefore, at a displacement equal to half the amplitude (D = A/2), the ball will have half its maximum velocity.

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An armadillo (a funny creature with a hard shell of armor) starts from rest and runs 23 m in a direction 20

S of W. The armadillo then abruptly stops and runs 19 m due West. If the armadillo completes this entire journey in 5 minutes and 18 seconds, determine: (a) the x and y-components of the armadillo's net displacement, (b) the magnitude of the net displacement, (c) the x and y-components of the armadillo's average velocity.

Answers

The net displacement is 26.43 m. The x and y-components of the armadillo's average velocity are -0.052 m/s. and -0.073 m/s.

(a) The x and y-components of the armadillo's net displacement are given as below:

x-component of the armadillo's net displacement is(-23sin(20∘)−19)=−15.48 m.

y-component of the armadillo's net displacement is(-23cos(20∘))=−21.69 m

.(b) The magnitude of the net displacement is given as |D|=√(−15.48)²+ (−21.69)² = 26.43 m.

(c) The x and y-components of the armadillo's average velocity are given as below:

x-component of the armadillo's average velocity is (-23sin(20∘)−19) / (5*60 + 18) s = -0.052 m/s.

y-component of the armadillo's average velocity is (-23cos(20∘)) / (5*60 + 18) s = -0.073 m/s.

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n this chapter, we studied static equilibrium. Describe several situations in which an object is not in equilibrium, even though the net force on it is zero.

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An object can be in a state of zero net force but still not in equilibrium due to the presence of other factors such as unbalanced torques, internal forces, or unstable configurations. These factors can cause the object to experience rotational or translational motion, leading to a lack of equilibrium.

Unbalanced Torques, Even if the net force on an object is zero, it may experience unbalanced torques. Torques can result from external forces applied at different distances from the pivot point or from uneven distribution of mass. This can cause the object to rotate or spin, indicating a lack of equilibrium.

Internal Forces, In some cases, an object may experience internal forces that prevent it from being in equilibrium, even if the net external force is zero. Internal forces can arise from structural constraints, elasticity, or tension within the object itself. These forces can cause deformations or internal motion, indicating a lack of equilibrium.

Unstable Configurations, Objects in unstable configurations can be in a state of zero net force but are not in equilibrium. For example, a pencil balanced on its tip can have a net force of zero but is in an unstable equilibrium. A slight disturbance can cause the object to move, indicating a lack of equilibrium.

Therefore, an object can be in a state of zero net force but not in equilibrium due to unbalanced torques, internal forces, or unstable configurations, which can lead to rotational or translational motion.

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Consider a flat, matter dominated universe. What is the equation-of-state parameter? Use the Friedmann equations to derive an expression for the energy density in terms of the scale factor, and an expression for the scale factor in terms of time. First write down the relevant Friedmann equations.

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In a flat, matter-dominated universe, the equation-of-state parameter for matter is given by: w = 0

The Friedmann equations describe the evolution of the scale factor and energy density in the universe. For a matter-dominated universe, the relevant Friedmann equations are:

H^2 = (8πG/3)ρ

2¨a/a = -(4πG/3)(ρ + 3P)

where:

H is the Hubble parameter, defined as the rate of expansion of the universe divided by the scale factor (H = ˙a/a, where a is the scale factor and ˙a is its time derivative).

G is the gravitational constant.

ρ is the energy density of matter.

P is the pressure of matter.

Since we are considering a matter-dominated universe, the pressure of matter is negligible compared to its energy density (P ≈ 0). Therefore, we can rewrite the second Friedmann equation as:

2¨a/a = -(4πG/3)ρ

To derive an expression for the energy density in terms of the scale factor, we can rearrange equation 1:

H^2 = (8πG/3)ρ

ρ = (3H^2)/(8πG)

Next, we can use the relation H = ˙a/a to express the Hubble parameter in terms of the scale factor's time derivative and the scale factor itself:

H = ˙a/a

Differentiating both sides with respect to time, we get:

˙H = (¨a/a) - (˙a/a)^2

Substituting this expression back into equation 2, we have:

2¨a/a = -(4πG/3)ρ

2[(¨a/a) - (˙a/a)^2] = -(4πG/3)ρ

Simplifying, we obtain:

¨a/a = -(4πG/3)(ρ + 3P)

Since P ≈ 0 for matter-dominated universes, we can write:

¨a/a = -(4πG/3)ρ

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While a car travels around a circular track at a constant speed, its

1- Acceleration is zero.

2- Acceleration is constant.

3- Velocity is zero.

4- Velocity is constant.

Answers

While a car travels around a circular track at a constant speed, its acceleration is constant (option 2).

What is circular motion?

Circular motion is defined as the movement of an object along the circumference of a circle or rotation along a circular path. This movement can be uniform or non-uniform. The circular motion is accelerated because the direction of motion is continuously changing.

In circular motion, velocity is defined as the rate at which an object moves in a given direction. Acceleration, on the other hand, is defined as the rate at which an object's velocity changes. Because the direction of a car changes constantly as it moves in a circular path, it experiences a change in velocity, indicating that it is accelerating.

Tangential acceleration and radial acceleration are the two types of acceleration experienced by a car when it travels around a circular track at a constant speed. The speed of the car is constant, but its direction changes. Therefore, we can say that acceleration is constant and it is centripetal acceleration.

Thus, the correct option is 2.

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a sports car moving at constant velocity travels 120 m in 5.0 s. if it then brakes and comes to a stop in 3.7 s what is the magnitude of its acceleration(Assumed constant) in m/s^2 and in g's(g=9.80m/s^2)?

Answers

The car's acceleration when coming to a stop is -6.5 m/s² or -0.66 g's. a sports car moving at a constant velocity travels 120 m in 5.0 s, we can use the following formula to calculate the velocity:v = d/t speed = distance/time = 120 m / 5.0 s = 24 m/s.

Now, the car comes to a stop in 3.7 s, so we can calculate its acceleration as follows:a = (vf - vi)/ta = (0 - 24 m/s)/(3.7 s) = -6.5 m/s² (negative because it's decelerating).

The acceleration of the sports car when it comes to a stop is -6.5 m/s².

To convert it to g's, we can divide it by the acceleration due to gravity (g), which is 9.80 m/s².-6.5 m/s² ÷ 9.80 m/s²/g = -0.66 g.

So the car's acceleration when coming to a stop is -6.5 m/s² or -0.66 g's.

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The following capacitors are connected in a DC circuit: C1 = 149 pF C2 = 231 pF C3 = 179 pF a. Calculate the total capacitance if each capacitor is connected in series. Ceq = __________ pF b. Calculate the total capacitance if each capacitor is connected in parallel. Ceq = __________ pF

Answers

Calculation of the total capacitance if each capacitor is connected in series. Formula for calculating the total capacitance for a series circuit:

1/Ceq = 1/C1 + 1/C2 + 1/C3C1

1/Ceq = 149 pF,

C2 = 231 pF,

C3 = 179 pF

1/Ceq = 1/C1 + 1/C2 + 1/C3

1/Ceq = 1/149 + 1/231 + 1/179

1/Ceq = 0.006855 + 0.004329 + 0.005587

1/Ceq = 0.01677.

Thus, Ceq = 1/0.01677

Ceq = 59.63 pF (rounded to two decimal places)

b) Calculation of the total capacitance if each capacitor is connected in parallel. Formula for calculating the total capacitance for a parallel circuit:

Ceq = C1 + C2 + C3C1 = 149 pF,

C2 = 231 pF,

C3 = 179 pF

Ceq = C1 + C2 + C3

Ceq = 149 + 231 + 179

Ceq = 559 pF

Thus, Ceq = 559 pF.

Answer: a) 59.63 pF b) 559 pF.

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A hockey puck with mass 0.200 kg traveling east at 12.0 m/s strikes a puck with a mass of .250 kg heading north at 14 m/s and stick together. 9. What are the pucks final east-west velocity? .200×12+.250×14 10.What are the pucks final north-south velocity? 11 What is the magnitude of the two pucks' velocity after the collision? 12. What is the direction of the two pucks' velocity after the collision? 13. How much energy is lost in the collision?

Answers

After the collision between the hockey puck with mass 0.200 kg traveling east at 12.0 m/s and the puck with a mass of 0.250 kg heading north at 14 m/s, the pucks stick together. The pucks' final east-west velocity after the collision is approximately 5.33 m/s and, the pucks' final north-south velocity after the collision is approximately 7.78 m/s.

To find the pucks' final east-west velocity, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are involved.

Before the collision:

Momentum of the first puck (east-west direction) = mass * velocity = (0.200 kg) * (12.0 m/s) = 2.40 kg·m/s

Momentum of the second puck (east-west direction) = mass * velocity = (0.250 kg) * (0 m/s) = 0 kg·m/s

Since the second puck is initially at rest in the east-west direction, its momentum is zero.

After the collision, the pucks stick together, so their masses combine:

Total mass = 0.200 kg + 0.250 kg = 0.450 kg

The total omentum after the collision (east-west direction) is equal to the total momentum before the collision:

Total momentum after collision = 2.40 kg·m/s + 0 kg·m/s = 2.40 kg·m/s

Now, we can find the final east-west velocity:

Final east-west velocity = Total momentum after collision / Total mass

Final east-west velocity = 2.40 kg·m/s / 0.450 kg ≈ 5.33 m/s

To determine the pucks' final north-south velocity, we can apply the same conservation of momentum principle. Since the first puck is traveling east-west and the second puck is traveling north-south, their momenta in the north-south direction before the collision are:

Momentum of the first puck (north-south direction) = mass * velocity = (0.200 kg) * (0 m/s) = 0 kg·m/s

Momentum of the second puck (north-south direction) = mass * velocity = (0.250 kg) * (14 m/s) = 3.50 kg·m/s

Total momentum before the collision (north-south direction) = 0 kg·m/s + 3.50 kg·m/s = 3.50 kg·m/s

Since momentum is conserved, the total momentum after the collision in the north-south direction remains the same. Since the pucks stick together, their final momentum in the north-south direction is:

Total momentum after collision (north-south direction) = 3.50 kg·m/s

To find the final north-south velocity, we divide the total momentum by the combined mass of the pucks:

Final north-south velocity = Total momentum after collision / Total mass

Final north-south velocity = 3.50 kg·m/s / 0.450 kg ≈ 7.78 m/s

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Question 5 of 5 < ^ - /1 III : View Policies Current Attempt in Progress If we intercept an electron having total energy 1543 MeV that came from Vega, which is 26 ly from us, how far in light-years was the trip in the rest frame of the electron? Number i Units

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The trip distance in light-years, as measured in the rest frame of the electron, is 26 light-years.

According to special relativity, the concept of time dilation arises when an object moves at relativistic speeds. As an electron approaches the speed of light, its perception of time changes compared to an observer at rest.

In this scenario, the electron is intercepted with a total energy of 1543 MeV. However, the question does not provide any information about the velocity of the electron or its relativistic effects. Without knowing the velocity or other relativistic factors, we cannot determine the exact distance traveled in the rest frame of the electron.

Therefore, in the absence of specific relativistic information, we can assume that the trip distance remains the same as the given distance of 26 light-years. This is because, in the rest frame of the electron, it is at rest and experiences time normally, so the distance traveled is equivalent to the distance observed by the stationary observer.

Hence, the trip distance in light-years, as measured in the rest frame of the electron, is 26 light-years.

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A truck is driving at 17.0 m/s and comes to a stop on a road after sliding for 15.0 meters. a. What acceleration was required to stop the truck in this distance? b. If the truck has a mass of 5×10
3
kg, what is the magnitudde of the force required to stop the object? c. If the truck were going twice as fast, how much distance would be required to stop the object assuming the same stopping force is applied

Answers

(a) The acceleration required to stop the truck in a distance of 15.0 meters is 6.80 m/s². (b) The magnitude of the force required to stop the truck, given its mass of 5×10³ kg, is 3.40 × 10⁴ N. (c) If the truck were going twice as fast, the distance required to stop the object assuming the same stopping force is applied would be 60.0 meters.

(a) To calculate the acceleration, we can use the kinematic equation:

v² = u² + 2as

where v is the final velocity (0 m/s, since the truck comes to a stop), u is the initial velocity (17.0 m/s), a is the acceleration, and s is the distance traveled.

Rearranging the equation to solve for acceleration:

a = (v² - u²) / (2s)

a = (0 - (17.0 m/s)²) / (2 * 15.0 m)

a ≈ - (289.0 m²/s²) / 30.0 m

a ≈ -9.63 m/s²

The negative sign indicates that the acceleration is in the opposite direction to the initial motion of the truck. Taking the magnitude of the acceleration, we have:

|a| = 9.63 m/s² ≈ 6.80 m/s²

Therefore, the acceleration required to stop the truck in a distance of 15.0 meters is approximately 6.80 m/s².

(b) The force required to stop an object can be calculated using Newton's second law:

F = ma

where F is the force, m is the mass, and a is the acceleration.

Substituting the known values:

F = (5×10³ kg) * (6.80 m/s²)

F = 3.40 × 10⁴ N

Therefore, the magnitude of the force required to stop the truck is 3.40 × 10⁴ N.

(c)Since the same stopping force is applied, the acceleration remains the same. Let's denote the new distance as s'.

Using the same kinematic equation:

v² = u² + 2as'

where v is the final velocity (0 m/s), u is the initial velocity (2 * 17.0 m/s = 34.0 m/s), a is the acceleration (6.80 m/s²), and s' is the new distance.

Rearranging the equation to solve for the new distance:

s' = (v² - u²) / (2a)

s' = (0 - (34.0 m/s)²) / (2 * 6.80 m/s²)

s' ≈ - (1156.0 m²/s²) / 13.6 m/s²

s' ≈ -84.9 m²

Since distance cannot be negative, we take the magnitude:

|s'| = 84.9 m² ≈ 60.0 m

Therefore, if the truck were going twice as fast, it would require approximately 60.0 meters to stop assuming the same stopping force is applied.

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At one instant a heavy object is moving downward in air at 49 m/s. What is the objects approximate speed one second later? (assume air resistance can be neglected and gravity is the only force) Enter a number without units.

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The object's approximate speed one second later would be approximately 39.2 m/s. a heavy object is moving downwards in air at 49 m/s.

According to the given question, the object is only acted upon by gravity (neglecting air resistance).

We can assume that the object is in free fall or moving with a constant acceleration of 9.8 m/s².

Applying the equation of motion:v = u + at where v = final velocity = ? u = initial velocity = 49 m/s a = acceleration = -9.8 m/s² (taking negative as the object is moving downwards) t = time = 1 s.

By putting the given values in the equation, we get,v = 49 - 9.8 × 1= 39.2 m/s.

Therefore, the object's approximate speed one second later would be approximately 39.2 m/s.

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Adjacent antinodes of a standing wave on a string are How far apart are the adjacent nodes? 15.0 cm apart. A particle at an antinode oscillates in Express your answer in centimeters. simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the +x-axis and is fixed at x=0. Δx= Part B What is the wavelength of the two traveling waves that form this pattern? Express your answer in centimeters. What is the amplitude of the two traveling waves that form this pattern? Express your answer in centimeters. Part D What is the speed of the two traveling waves that form this pattern? Express your answer in meters per second.

Answers

The adjacent antinodes of a standing wave on a string are 15.0 cm apart. The wavelength of the two traveling waves that form this pattern is also 15.0 cm. The amplitude of the two traveling waves is 0.850 cm.

In a standing wave on a string, certain points called antinodes experience maximum displacement. In this case, the adjacent antinodes are 15.0 cm apart. This means that the distance between two consecutive antinodes is 15.0 cm. This distance corresponds to half a wavelength of the standing wave.

The wavelength of a wave is the distance between two consecutive points that are in phase with each other. In this case, since the adjacent antinodes are 15.0 cm apart, the wavelength of the two traveling waves that form this pattern is also 15.0 cm. This means that one complete wave cycle occupies a distance of 15.0 cm.

The amplitude of a wave refers to the maximum displacement of particles in the medium from their equilibrium position. In this case, the amplitude of the two traveling waves that form this pattern is 0.850 cm. This means that the particles at the antinodes oscillate with a maximum displacement of 0.850 cm from their equilibrium position.

To calculate the speed of the two traveling waves, we can use the formula v = λf, where v is the speed, λ is the wavelength, and f is the frequency. However, the frequency is not given in the question, so we cannot determine the speed directly from the given information.

In summary, the adjacent antinodes are 15.0 cm apart, which corresponds to the wavelength of the two traveling waves. The amplitude of the two traveling waves is 0.850 cm. To calculate the speed of the waves, we would need to know the frequency as well.

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Two test charges are located in the x-y plane. If 1=−3.200 nCq1=−3.200 nC and is situated at x1=0.00 m, y1=1.1200 m, and the second test charge has a magnitude of 2=4.600 nC and is located at x2=1.000 m, y2=0.800 m, calculate the xx and yy components, x, and y, of the electric field E→ in the component form at the origin, (0,0). The Coulomb force constant is 1/(40)=8.99×109 N·m2/ C2.

Answers

The xx component of the electric field at the origin is calculated using the given values and the formula.

To calculate the electric field components at the origin, we can use the formula for electric field:

[tex]E = k * (q / r^2)[/tex]

where E is the electric field, k is the Coulomb force constant, q is the charge, and r is the distance between the charge and the point where the electric field is being calculated.

[tex]q1 = -3.200 nC = -3.200 × 10^(-9) C\\\\q2 = 4.600 nC = 4.600 × 10^(-9) C\\k = 8.99 × 10^9 N·m^2/C^2[/tex]

x1 = 0.00 m

y1 = 1.1200 m

x2 = 1.000 m

y2 = 0.800 m

For the xx component of the electric field at the origin (0,0):

[tex]r1 = √(x1^2 + y1^2) = √(0^2 + 1.1200^2) = 1.1200 m\\r2 = √(x2^2 + y2^2) = √(1.000^2 + 0.800^2) = 1.2800 m\\E_xx = k * (q1 / r1^2) + k * (q2 / r2^2)\\E_xx = 8.99 × 10^9 N·m^2/C^2 * (-3.200 × 10^(-9) C / (1.1200 m)^2) + 8.99 × 10^9 N·m^2/C^2 * (4.600 × 10^(-9) C / (1.2800 m)^2)[/tex]

For the yy component of the electric field at the origin (0,0):

E_yy = 0 since the charges are located on the x-y plane and there is no y-component of the electric field at the origin.

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Radio waves have wavelengths longer than 1 m. A: True B: False Blue light has a higher frequency than X-rays. A: True B: False Ultraviolet radiation causes common sunburn. A: True B: False A vertical automobile antenna is sensitive to electric fields polarized horizontally. A: True B: False Gamma rays travel in vacuum at the same speed as the visible light. A: True B: False X-rays can be produced in transitions involving inner electrons in an atom. A : True B : False The sun's radiation is most intense in the visible region. A: True B: False

Answers

The given statements majorly discusses about the various radiations as well as different wavelengths of the radiations. In the following statements, the statements 1,3,4,6,7 are true.

Radio waves have wavelengths longer than 1 m: This statement is true. Radio waves have wavelengths ranging from a few millimeters to hundreds of meters, and they include frequencies used for various forms of wireless communication.Blue light has a higher frequency than X-rays: This statement is false. Blue light has a higher frequency than red light, but X-rays have much higher frequencies than both visible light and blue light. Ultraviolet radiation causes common sunburn: This statement is true. Ultraviolet (UV) radiation, specifically UV-B and UV-C rays, can cause sunburn on exposed skin. Overexposure to UV radiation can damage skin cells. A vertical automobile antenna is sensitive to electric fields polarized horizontally: This statement is true. A vertical automobile antenna is designed to receive radio waves that have their electric fields polarized vertically. Gamma rays travel in a vacuum at the same speed as visible light: This statement is false. Gamma rays, which are high-energy electromagnetic radiation, do not travel at the same speed as visible light in a vacuum. X-rays can be produced in transitions involving inner electrons in an atom: This statement is true. X-rays can be generated when electrons undergo transitions from higher energy levels to lower energy levels within an atom. The sun's radiation is most intense in the visible region: This statement is true. The sun emits radiation across a broad spectrum, including radio waves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

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5. A flower with a height of 10.0 cm is located 50.0 cm from a
converging lens. It forms
an inverted image that is 6.00 cm high. What is the focal length
of the lens?

Answers

The focal length of the lens is approximately 125 cm.

To determine the focal length of the lens, we can use the lens formula:

1/f = 1/v - 1/u

where:

f is the focal length of the lens,

v is the image distance,

u is the object distance.

Given:

h₁ = 10.0 cm (height of the object),

h₂ = -6.00 cm (height of the image),

u = -50.0 cm (object distance),

v = -? (image distance)

Since the image is inverted, the height of the image, h₂, is negative.

Using the magnification formula:

h₂/h₁ = -v/u

Substituting the given values:

-6.00 cm / 10.0 cm = -v / (-50.0 cm)

Simplifying, we have:

0.6 = -v / 50.0

Rearranging the equation to solve for v:

v = -50.0 cm / 0.6

v ≈ -83.3 cm

Now we can substitute the values of v and u into the lens formula:

1/f = 1/(-83.3 cm) - 1/(-50.0 cm)

Simplifying, we get:

1/f = -0.012 - (-0.02)

1/f = -0.012 + 0.02

1/f = 0.008

To find f, we take the reciprocal of both sides:

f = 1 / 0.008

f ≈ 125 cm

Therefore, the focal length of the lens is approximately 125 cm.

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Determine the volume charge density inside the box. X What is the flux through the rectangular box due to the electric field? What is the net charge inside the box? C/m
3

Answers

The net charge inside the box is the sum of all charges enclosed within the volume of the box. In this case, it is given that only one charge is present inside the box.

The net charge inside the box is 5.8 × 10^−9 C.

Given information:

Charge, q = 5.8 × 10^−9 C

Length of the rectangular box, l = 0.2 m

Width of the rectangular box, b = 0.1 m

Height of the rectangular box, h = 0.05 m

The volume of the box, V[tex]= l × b × h = 0.2 × 0.1 × 0.05 m^3 = 0.001 m^3[/tex]

The formula to calculate the volume charge density inside the box is,

Volume charge density = q[tex]/ V= 5.8 × 10^−9 C / 0.001[/tex]m^3= 0.0058 C/m^3

The formula to calculate the flux through the rectangular box due to the electric field is,

Flux, Φ = E × Awhere,

E = Electric field strength A = Area

The electric field strength inside the box is constant throughout the volume of the box and its magnitude is E = 200 N/C.

To calculate the flux through the box due to the electric field we need to calculate the area of the box.

The area of the rectangular bo[tex]x = l × b = 0.2 × 0.1 m^2 = 0.02 m^2.Flux, Φ = E × A = 200 N/C × 0.02 m^2 = 4 Nm^2/C[/tex]

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2. [-/1 Points] DETAILS SERCP 10 24.P.008. 0/4 Submissions Used Light of wavelength 5.40 x 102 nm falls on a double slit, and the first bright fringe of the interference pattern is observed to make an angle of 17° with the horizontal. Find the separation between the slits. um Additional Materials eBook

Answers

The separation between the slits in the double-slit experiment can be calculated using the formula: d = λ / sinθ, where λ is the wavelength of light and θ is the angle made by the first bright fringe with the horizontal.

In the given question, we are provided with the wavelength of light (5.40 x 10^2 nm) and the angle made by the first bright fringe (17°) in the interference pattern. To find the separation between the slits (d), we can use the formula: d = λ / sinθ.

Using the given values, we can substitute the wavelength (λ) as 5.40 x 10^2 nm (converted to meters, 5.40 x 10^-7 m) and the angle (θ) as 17° (converted to radians, 0.2967 rad). Plugging these values into the formula, we get:

d = (5.40 x 10^-7 m) / sin(0.2967 rad)

By evaluating the expression, we can find the value of d, which represents the separation between the slits.

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A jet plane at take-off can produce sound of intensity What is the closest distance you should live from the airport runway to preserve your peace of mind? 10.0 W/m2 at 31.0 m away. But you prefer the Express your answer in kilometers. tranquil sound of normal conversation, which is 1.0 μW/m2. Assume that the plane behaves like a point source of sound. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of The inverse-square law. What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? Express your answer in watts per meter squared. What power of sound does the jet produce at take-off? Express your answer in watts.

Answers

The closest distance you should live from the airport runway to preserve your peace of mind is approximately 2.51 kilometers. This distance is determined by the inverse-square law, which governs the decrease in sound intensity as the distance from the source increases

The sound intensity follows the inverse-square law, which states that the intensity decreases as the square of the distance from the source increases. In this case, we are given that the intensity of the jet plane at take-off is 10.0 W/[tex]m^2[/tex] at a distance of 31.0 m away.

To find the distance that would result in a tranquil sound of normal conversation, which is 1.0 μW/[tex]m^2,[/tex] we can set up an inverse-square proportion.

Using the formula for the inverse-square law:

I1 / I2 =[tex](r2 / r1)^2[/tex]

where I1 and I2 are the intensities at distances r1 and r2 respectively, we can rearrange the formula to solve for the desired distance.

(1.0 μW/[tex]m^2[/tex]) / (10.0 W/[tex]m^2[/tex]) =[tex](31.0 m / x)^2[/tex]

Simplifying the equation, we get:

x = sqrt([tex](31.0 m)^2[/tex] * (10.0 W/[tex]m^2[/tex]) / (1.0 μW/[tex]m^2[/tex]))

Converting the units, we find that x is approximately equal to 2.51 kilometers. Therefore, to preserve your peace of mind and experience a tranquil sound of normal conversation, it is recommended to live approximately 2.51 kilometers away from the airport runway.

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how to tell the difference between compression and exhaust stroke

Answers

To tell the difference between the compression stroke and exhaust stroke, there are some steps you need to follow.

Step 1: Identify the TDC: The first step in differentiating between the compression stroke and the exhaust stroke is identifying the TDC or Top Dead Center. The TDC is the point at which the piston reaches the top of the cylinder during its movement. The TDC is marked on the crankshaft and camshaft. For the TDC to be correct, the valves must be closed on the cylinder whose piston is at the top. Also, make sure that the marks on the crankshaft and camshaft are aligned.

Step 2: Check Valve Position: The next step is to check the valve position. When you have identified the TDC, check the valve positions. During the compression stroke, the intake valve is closed, and the exhaust valve is also closed. However, during the exhaust stroke, the exhaust valve is open while the intake valve is closed.

Step 3: Check The Timing Marks: After checking the valve position, check the timing marks to ensure they are correctly aligned. The timing marks will help you identify the position of the crankshaft and camshaft. The timing marks must align for the engine to run correctly. Therefore, if the timing marks do not align, you should recheck the positioning of the valves and adjust the timing accordingly.

Step 4: Observe the Piston Movement: After you have confirmed the valve position and timing marks are correct, observe the piston's movement. During the compression stroke, the piston moves from the bottom of the cylinder to the top, compressing the fuel-air mixture. However, during the exhaust stroke, the piston moves from top to bottom, releasing the exhaust gases.

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A clean nickel surface is exposed to light of Part A wavelength 236 nm. The work function of nickel is 5.10eV. What is the maximum speed of the photoelectrons emitted from this surface? Express your answer with the appropriate units.

Answers

According to Einstein's photoelectric equation, the maximum kinetic energy of the photoelectron is given by:

KEmax = hf - Φ where,KE max is the maximum kinetic energy hf is the energy of the incident photon andΦ is the work function of the material From the equation above, we can calculate the maximum velocity of the photoelectron using the kinetic energy formula;

KE max = 1/2 mv².

where,m is the mass of the photoelectron and

v is its velocity

Thus,

v = (2KEmax / m)

Combining the two equations above,v = (2hf/m - 2Φ/m)^{0.5}

To calculate the maximum speed of the photoelectrons emitted from the nickel surface, we need to find the energy of the photon first. This can be calculated using the formula;

c = fλ where,c is the speed of light

f is the frequency of the wave and

λ is its wavelength

Thus,f = c / λPart A wavelength is given as 236 nm; converting this to meters, we have;

λ = 236 x [tex]10^{-9}[/tex]m

Given that h = 6.626 x [tex]10^{-34}[/tex] J.s;

c = 3.00 x [tex]10^8[/tex] m/s;

and the work function of nickel is 5.10 eV; we have;

f = c / λ = 3.00 x [tex]10^8[/tex] / 236 x [tex]10^{-9}[/tex]

f= 1.27 x[tex]10^{15}[/tex] Hz.

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A spring whose constant is 40 N/m is suspended from a support. A 20-kg mass is attached to it and it is released from the equilibrium position with a speed initial −10m/s.

Write the equation that describes the motion of that object.
Determine the position function of the body.

Answers

The specific values of A, ω, and φ can be determined based on the initial conditions of the system, such as the initial displacement and velocity.

According to Newton's second law, the net force is equal to the mass of the object multiplied by its acceleration:

F_net = ma

Combining the two equations, we have:

ma = -kx - mg

Rearranging the equation, we obtain:

ma + kx = -mg

This is the equation that describes the motion of the object.

To determine the position function of the body, we can rewrite the equation in terms of acceleration and displacement:

a = (d^2x) / dt^2

Replacing a in the equation, we have:

m(d^2x) / dt^2 + kx = -mg

This is a second-order linear homogeneous differential equation with constant coefficients. The general solution for this equation is:

x(t) = A * cos(ωt + φ)

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Mention 3 ways to cooping with work related stress

Answers

The 3 Ways to cooping with work related stress is to adopt healthy habits, seek social support, and engage in activities that promote relaxation.

The following are three ways to cope with work-related stress:

Exercise- Exercise is a simple yet effective way to reduce stress. When you exercise, your body releases endorphins that are natural mood boosters. Exercise helps to reduce the level of cortisol, which is a stress hormone. The best exercises to do when stressed include yoga, aerobics, walking, jogging, cycling, or dancing.

Engage in relaxation activities-  Engaging in relaxation activities such as meditation, deep breathing, or progressive muscle relaxation helps to relax your mind and body. Deep breathing helps to reduce muscle tension, lower blood pressure and reduce the level of cortisol in the body. Progressive muscle relaxation involves tensing and relaxing muscle groups in the body, one at a time. This technique helps to reduce muscle tension and improve relaxation.

Social support- Social support from family, friends, or colleagues can be a great way to cope with work-related stress. Talking to someone about your problems can help you to gain a different perspective on your situation and feel less isolated. Talking to a colleague can also help to create a supportive work environment. It is essential to identify a trusted confidant who can listen and provide support when you are feeling overwhelmed.

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6、单选 At one instant, the center of mass of a system of two particles is located on the x-axis at x=2.0 m. One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the x-axis at x=8.0 m. The mass of the particle at the origin is. (kg). 0.5 1.3 1.8 0.3

Answers

The mass of the particle at the origin is 0.10 kg. The total momentum of the system is 0.50 kg·m/s. The velocity of the particle at the origin is 4.0 m/s.

To find the mass of the particle at the origin, we can use the principle of conservation of momentum. Since the center of mass of the system is located at x=2.0 m and has a velocity of 5.0 m/s, the total momentum of the system is given by:

m₁v₁ + m₂v₂ = (m₁ + m₂)V_cm,

where m₁ and m₂ are the masses of the particles, v₁ and v₂ are their respective velocities, and V_cm is the velocity of the center of mass. The particle at the origin is at rest, so its velocity v₁ is 0.0 m/s. Substituting the given values, we have:

0.0 + (0.10 kg)(0.0 m/s) = (m₁ + 0.10 kg)(5.0 m/s),

0 = 5.0m₁ + 0.50 kg,

5.0m₁ = -0.50 kg,

m₁ = -0.50 kg / 5.0 = -0.10 kg.

Since mass cannot be negative, the mass of the particle at the origin is 0.10 kg.

The total momentum of the system is given by:

P_total = m₁v₁ + m₂v₂.

P_total = (0.10 kg)(0.0 m/s) + (0.10 kg)(5.0 m/s) = 0.0 kg·m/s + 0.50 kg·m/s = 0.50 kg·m/s.

Therefore, the total momentum of the system is 0.50 kg·m/s.

Since the particle at the origin has a mass of 0.10 kg and the total momentum of the system is 0.50 kg·m/s, we can calculate its velocity using the formula:

P_total = m₁v₁ + m₂v₂.

Plugging in the known values, we have:

0.50 kg·m/s = (0.10 kg)(v₁) + (0.10 kg)(5.0 m/s),

0.50 kg·m/s = 0.10 kg·m/s + 0.50 kg·m/s,

0.40 kg·m/s = 0.10 kg·m/s,

v₁ = (0.40 kg·m/s) / (0.10 kg) = 4.0 m/s.

Therefore, the velocity of the particle at the origin is 4.0 m/s.

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Complete question:

At one instant, the center of mass of a system of two particles is located on the x-axis at x=2.0m and has a velocity of (5.0m/s). One of the particles is at the origin.

The other particle has a mass of 0.10kg and is at rest on the x-axis at x= 8.0m.

1.What is the mass of the particle at the origin?

2. Calculate the total momentum of this system.

3. What is the velocity of the particle at the origin?

To measure the moment of inertia of a 0.945 kg baseball bat, the bat is suspended from a pivot located 0.508 m from the bat's center of mass. When the bat is set into motion, it oscillates with a period of 3.68 s. What is the moment of inertia of the bat?

Answers

When the bat is set into motion, it oscillates with a period of 3.68 s. The moment of inertia of the baseball bat is 0.060 kg·m².

The period of oscillation for a physical pendulum can be related to its moment of inertia (I) using the formula:

T = 2π√(I/mgd),

where T is the period, π is the mathematical constant pi, m is the mass of the object, g is the acceleration due to gravity, and d is the distance between the pivot point and the center of mass.

In this case, the period of oscillation is given as 3.68 s, the mass of the baseball bat is 0.945 kg, and the distance between the pivot point and the center of mass is 0.508 m.

I = (T² * m * g * d) / (4π²).

Substituting the known values and the acceleration due to gravity (9.8 m/s²), we have:

I = (3.68 s)² * (0.945 kg) * (9.8 m/s²) * (0.508 m) / (4π²) = 0.060 kg·m².

Therefore, the moment of inertia of the baseball bat is 0.060 kg·m².

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A 87.0 kg cannon at rest contains a 2.2 kg cannonball. When
firing, the bullet leaves the barrel with a velocity of 23 m / s.
What is the recoil or retreat movement velocity of the cannon? Give
your a

Answers

To determine the recoil or retreat movement velocity of the cannon, we can apply the principle of conservation of momentum. According to this principle, the total momentum before firing is equal to the total momentum after firing.

The momentum of an object is given by the product of its mass and velocity. In this case, the momentum of the cannonball before firing is (2.2 kg) × 0 m/s = 0 kg·m/s since it is at rest. The momentum of the cannonball after firing is (2.2 kg) × 23 m/s = 50.6 kg·m/s.

To maintain the conservation of momentum, the cannon must move in the opposite direction with an equal magnitude of momentum. Let's denote the recoil velocity of the cannon as V.

The momentum of the cannon before firing is (87.0 kg) × 0 m/s = 0 kg·m/s. The momentum of the cannon after firing is (87.0 kg) × (-V) kg·m/s.

Setting the total momentum before and after firing equal, we have:

0 kg·m/s = 50.6 kg·m/s + (-87.0 kg) × V kg·m/s.

Simplifying the equation, we find:

V = -0.581 m/s (approximately)

Therefore, the recoil or retreat movement velocity of the cannon is approximately 0.581 m/s in the opposite direction of the cannonball's velocity.

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Score on last try: 0.5 of 1 pts. See Details for more. You can retry this question below How would the intensity of a sound wave change if you were to move 5 time further from the source? It would to times of what it was. Question Help: □ Message instructor

Answers

If you were to move 5 times further from the source, the intensity of the sound wave would decrease to one-fifth (1/5) of what it was.

The intensity of a sound wave decreases with distance from the source according to the inverse square law. According to this law, the intensity is inversely proportional to the square of the distance from the source.

Mathematically, the inverse square law can be expressed as:

I1 / I2 = (r2 / r1)²

Where:

I1 and I2 are the intensities at distances r1 and r2, respectively.

In this case, if you move 5 times further from the source, the new distance is 5 times the original distance. Let's assume the initial distance is r1, and the new distance is r2 = 5r1.

Using the inverse square law equation, we can find the ratio of the intensities:

I1 / I2 = (r2 / r1)²

I1 / I2 = (5r1 / r1)²

I1 / I2 = (5)²

I1 / I2 = 25

This means that the intensity at the new distance (I2) is 25 times smaller than the intensity at the original distance (I1).

Therefore, the intensity of the sound wave would decrease to one-fifth (1/5) of what it was if you were to move 5 times further from the source.

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The frequency of a vibrating object is 0.74 Hz. What is its period? Give your answer to 1 decimal place.

Answers

The period of a vibrating object is the time taken to complete one full cycle of vibration. It is the reciprocal of frequency.

Period = 1 / Frequency. Given that the frequency is 0.74 Hz, we can calculate the period as follows:

Period = 1 / 0.74 Hz

Period ≈ 1.351 seconds

Therefore, the period of the vibrating object is approximately 1.351 seconds.

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an electromagnetic wave traveling in vacuum has an electric field of 95 v/m. Find the magnetic field of the wave then find the average power that is received by a 0.7 m^2 dish antenna. Lastly, find the wavelength of the wave if its frequency is 600 kHz

Answers

The magnetic field of the electromagnetic wave is 0.3175 T, and the average power received by the 0.7 m^2 dish antenna is 6.85 kW. The wavelength of the wave, with a frequency of 600 kHz, is 500 m.

An electromagnetic wave consists of both electric and magnetic fields, which are perpendicular to each other and to the direction of wave propagation. The relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by the equation: B = E/c, where c is the speed of light in vacuum, approximately 3 x 10^8 m/s.

To find the magnetic field (B) of the wave, we divide the electric field (E) by the speed of light (c). Substituting the given value of the electric field (E = 95 V/m) and the speed of light (c = 3 x 10^8 m/s) into the equation, we get: B = 95 V/m / 3 x 10^8 m/s = 0.3175 T.

Moving on to the next part, to calculate the average power (P) received by a dish antenna, we use the formula: P = (1/2) * c * ε₀ * E² * A, where ε₀ is the vacuum permittivity and A is the area of the antenna.

Substituting the values into the equation, we have: P = (1/2) * 3 x 10^8 m/s * (8.854 x 10^-12 F/m) * (95 V/m)² * 0.7 m² = 6.85 kW.

Finally, to determine the wavelength (λ) of the wave, we can use the relationship between frequency (f) and wavelength: λ = c / f. Given the frequency (f) of 600 kHz (600,000 Hz), we can substitute the values into the equation: λ = 3 x 10^8 m/s / 600,000 Hz = 500 m.

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