a projectile has initial speed 0, angle . what is its ke when it is on the way down at a height ℎabove the ground? (assume no air resistance.)

Answers

Answer 1

Its kinetic energy at this point can be obtained by using the equation:KE = 1/2mv² = 1/2m(v₀sinθ-gt)²Thus, the kinetic energy of the projectile when it is on the way down at a height ℎ above the ground can be calculated using the formula KE = 1/2m(v₀sinθ-gt)².

A projectile with an initial speed 0 and angle can attain kinetic energy when it is moving. When the projectile is in the way down, and it is ℎabove the ground, it can also have kinetic energy. The formula for kinetic energy is KE

= 1/2mv² where m is mass, v is velocity, and KE is kinetic energy.What is kinetic energy.Kinetic energy is the energy that a moving body possesses. The amount of energy is equal to one-half the mass of the object and the square of its velocity. Thus, an object with a greater mass and speed will have more kinetic energy than a smaller object with a lower speed.Content loaded projectile If a content-loaded projectile has an initial speed of 0 and an angle of release θ with respect to the horizontal, its velocity at any point in time is given by:v

= v₀cosθî + (v₀sinθ-gt)ĵ

Where:v₀ is the initial speedθ is the angle of release g is the acceleration due to gravity is the time taken from release In the case of a projectile that is ℎ above the ground, and assuming there is no air resistance, the potential energy is given by mgh. When the projectile is in the way down, the KE formula applies, KE

= 1/2mv², but the velocity in this case is the vertical component of the projectile's velocity when it hits the ground.The vertical component of the velocity when the projectile is in the way down is given by:v

= v₀sinθ - gt

When the projectile is in the way down and is at a height ℎ above the ground, its potential energy is given by mgh. Its kinetic energy at this point can be obtained by using the equation:KE

= 1/2mv²

= 1/2m(v₀sinθ-gt)²

Thus, the kinetic energy of the projectile when it is on the way down at a height ℎ above the ground can be calculated using the formula KE

= 1/2m(v₀sinθ-gt)².

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Related Questions

a group of friends gather around a fire to stay warm. this is an example of what type of heat transfer? question 4 options: radiation convection conduction open system

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The example of friends gathering around a fire to stay warm is an example of heat transfer through radiation.

In this scenario, the heat from the fire is emitted in the form of electromagnetic radiation (infrared), which travels through the space and is absorbed by the people and objects nearby.

The transfer of heat occurs without direct contact or the need for a medium to carry the heat.

Hence, The example of friends gathering around a fire to stay warm is an example of heat transfer through radiation.

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A plane lands on a runway with a speed of 105 m/s, moving east, and it slows to a stop in 15.0 s. What is the magnitude (in m/s2) and direction of the plane's average acceleration during this time interval

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The magnitude of the plane's average acceleration during this time interval is 7 m/s², and its direction is west.

To determine the magnitude of average acceleration, we can use the formula:

Average Acceleration = (Change in Velocity) / (Time Interval)

The change in velocity can be calculated by subtracting the final velocity from the initial velocity:

Change in Velocity = Final Velocity - Initial Velocity

Change in Velocity = 0 m/s - 105 m/s = -105 m/s

Since the plane is slowing down, the change in velocity is negative. Therefore, the magnitude of the average acceleration is given by:

Magnitude of Average Acceleration = |-105 m/s| / 15.0 s = 7 m/s²

The negative sign indicates that the plane's velocity is decreasing, and its direction of motion is opposite to its initial direction. Since the plane was initially moving east, the direction of the average acceleration is west.

Thus, the magnitude of the plane's average acceleration during this time interval is 7 m/s², and its direction is west.

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A cook heats 500 g of olive oil in a steel pan which has a mass of 300 g. The oil needs to be heated from 20 °C to 190 °C. Using data from Table 10.1, calculate the thermal energy needed:

a to heat the pan
b to heat the oil
c the total

Answers

a. The thermal energy needed to heat the pan is 25,500 Joules.

b. The thermal energy needed to heat the oil is 153,000 Joules.

c.  The total thermal energy needed to heat both the pan and the oil is 178,500 Joules.

To calculate the thermal energy needed to heat the pan and the oil, we can use the equation:

Q = mcΔT,

where

Q = thermal energy

m = mass

c = specific heat capacity

ΔT = change in temperature.

First, let's calculate the thermal energy needed to heat the pan:

a) Heating the pan:

Given:

Mass of the pan (m1) = 300 g = 0.3 kg

Specific heat capacity of steel (c1) = 500 J/kg°C (from Table 10.1)

Change in temperature (ΔT1) = 190 °C - 20 °C = 170 °C

Q1 = m1c1ΔT1

= (0.3 kg)(500 J/kg°C)(170 °C)

= 25,500 J

Therefore, the thermal energy needed to heat the pan is 25,500 Joules.

b) Heating the oil:

Given:

Mass of the oil (m2) = 500 g = 0.5 kg

Specific heat capacity of olive oil (c2) = 1,800 J/kg°C (from Table 10.1)

Change in temperature (ΔT2) = 190 °C - 20 °C = 170 °C

Q2 = m2c2ΔT2

= (0.5 kg)(1800 J/kg°C)(170 °C)

= 153,000 J

Therefore, the thermal energy needed to heat the oil is 153,000 Joules.

c) Total thermal energy:

To find the total thermal energy, we sum up the thermal energies for heating the pan and the oil:

Total thermal energy (Qtotal) = Q1 + Q2

= 25,500 J + 153,000 J

= 178,500 J

Therefore, the total thermal energy needed to heat both the pan and the oil is 178,500 Joules.

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in nec 210.52(a)(1), the "6 foot rule" for spacing receptacles applies to all the following areas of a house, except for ____

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In NEC 210.52 (a) (1),  the "6 foot rule" for spacing receptacles applies to all areas of a house, except for bathrooms.

The "6 foot rule" stated in NEC 210.52 (a) (1) requires that there should be no more than 6 feet of unbroken wall space between receptacles in dwelling units. This rule ensures that electrical outlets are conveniently placed throughout a home to provide easy access to power sources. However, bathrooms have different requirements for receptacle spacing due to safety considerations.

NEC 210.52 (d) specifies that at least one receptacle outlet must be installed within 3 feet of the outside edge of each basin or sink in a bathroom. This rule aims to minimize the use of extension cords and potential electrical hazards in wet areas. To summarize, the "6 foot rule" for spacing receptacles applies to all areas of a house, except for bathrooms. Bathrooms have their own specific requirements for receptacle placement to ensure safety in potentially wet environments.

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Coherent light with wavelength 200 nm passes through two identical slits. The width of each slit is a, and the distance between the centers of the slits is d=1.00 mm. The m= 5 maximum in the two-slit interference pattern is absent, but the maxima for m= 0 through m= 4 are present Part A What is the ratio of the intensities for the m= 1 and m 2 maxima in the two-slit pattern? Express your answer using three significant figures.

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Coherent light with wavelength 200 nm passes through two identical slits. The width of each slit is a, and the distance between the centers of the slits is d=1.00 mm. The m= 5 maximum in the two-slit interference pattern is absent, but the maxima for m= 0 through m= 4 are present, the ratio of the intensities for the m = 1 and m = 2 maxima in the two-slit interference pattern is approximately 0.554

In a two-slit interference pattern, the intensity at a particular maximum is given by:

I = I₀ × cos²(θ)

where I₀ is the intensity of the central maximum, and θ is the angle from the central maximum to the specific maximum.

The angle θ can be calculated using the formula:

θ = m × λ / d

where m is the order of the maximum, λ is the wavelength of light, and d is the distance between the centers of the slits.

Given:

Wavelength, λ = 200 nm = 200 × 10^(-9) m

Distance between slits, d = 1.00 mm = 1.00 × 10^(-3) m

We are interested in finding the ratio of the intensities for the m = 1 and m = 2 maxima. So we need to calculate the values of I₁ and I₂ using the above equations.

For m = 1:

θ₁ = (1 × λ) / d

For m = 2:

θ₂ = (2 × λ) / d

Now let's calculate the intensity ratio:

I₁ / I₂ = (I₀ × cos²(θ₁)) / (I₀ × cos²(θ₂))

= cos²(θ₁) / cos²(θ₂)

Substituting the values of θ₁ and θ₂, we have:

I₁ / I₂ = cos²((λ / d) / cos²((2λ / d))

I₁ / I₂ = cos²((200 × 10^(-9)) / (1.00 × 10^(-3))) / cos²((2 × 200 × 10^(-9)) / (1.00 × 10^(-3)))

Using a calculator, we can evaluate the ratio:

I₁ / I₂ ≈ 0.554

Therefore, the ratio of the intensities for the m = 1 and m = 2 maxima in the two-slit interference pattern is approximately 0.554 (rounded to three significant figures).

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. Enter your answer in the provided box. 1. What is the equilibrium constant for the acid-base reaction between ammonia and acetic acid? Keq = 10
2. Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid?
Acid pKa
Hydrobromic Acid −5.8
Ethyloxonium Ion −2.4

Answers

"The equilibrium constant is temperature-dependent, meaning that the value of Keq can change with changes in temperature." The equilibrium constant (Keq) is a quantitative measure of the position of a chemical equilibrium for a given reaction.

It represents the ratio of the concentrations (or partial pressures for gas-phase reactions) of the products to the concentrations (or partial pressures) of the reactants, with each concentration term raised to the power of its stoichiometric coefficient. To determine the equilibrium constant (Keq) for the acid-base reaction between ethanol and hydrobromic acid, we need the pKa values of both species involved. However, you have provided the pKa values for hydrobromic acid and the ethyl oxonium ion. Ethanol itself does not have a pKa value since it is not an acid.

For a generic chemical reaction:

aA + bB ⇌ cC + dD

The equilibrium constant, Keq, is expressed as:

Keq = [C]c [D]d / [A]ᵃ [B]ᵇ

where [A], [B], [C], and [D] represent the molar concentrations of the species A, B, C, and D, respectively.

The value of Keq indicates the extent to which the reaction favors the formation of products (Keq > 1) or reactants (Keq < 1) at equilibrium. A Keq value of exactly 1 signifies that the concentrations of reactants and products are equal at equilibrium, indicating a balanced reaction.

The equilibrium constant is temperature-dependent, meaning that the value of Keq can change with changes in temperature.

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Free neutrons have a characteristic half-life of 10.4 min. What fraction of a group of free neutrons with kinetic energy 0.0400 eV decays before traveling a distance of 10.0 km ?

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The fraction of neutrons decayed before traveling a distance of 10.0 km is 0.004.

The half-life of free neutrons, t1/2 = 10.4 min

The initial kinetic energy of free neutrons, E = 0.0400 eV

The distance traveled by neutrons, d = 10.0 km

We know that the half-life of a radioactive substance is the time in which half of the substance decays or disintegrates.

Let N₀ be the number of neutrons at the beginning, then the number of neutrons N after time t can be given as:

N = N₀ / 2^(t / t1/2)

Here, t is the time and t1/2 is the half-life of the neutrons.

Initially, the number of neutrons N₀ is equal to 1.

Since we want to find the fraction of neutrons decayed, we can use the number of neutrons remaining at distance d from the source of neutrons.

Let's find the number of neutrons that decayed after traveling a distance of d = 10.0 km.

The speed of free neutrons is given as v = √(2E / m)

where m is the mass of the neutron.

Neutrons with kinetic energy E = 0.0400 eV will have a speed of v = √(2 * 0.0400 eV / 1.675 x 10⁻²⁷ kg) = 2.76x 10³ m/s

(0.04 eV = 6.409 × 10⁻²¹ Joule)

The time taken to travel a distance of 10.0 km is given as:

t = d / v = 10.0 x 10³ m / 2.76 x 10³ m/s = 3.6 s

Now, the number of neutrons remaining N' after a time t is:

N' = N₀ / 2^(t / t1/2)

Putting the values, we get: N' = 1 / 2^(3.6 s / 10.4 min) = 0.996             (10.4 min = 624 s)

The fraction of neutrons decayed is given as f = (N₀ - N') / N₀ = (1 - 0.996) / 1 = 0.004

Therefore, the fraction of neutrons decayed before traveling a distance of 10.0 km is 0.004.

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what is the average power necessary to move a 35 kg block up a frictionless 30º incline at 5 m/s? group of answer choices 68 w 121 w 343 w 430 w 860 w

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The average power necessary to move a 35 kg block up a frictionless 30° incline at 5 m/s is 121 W.

To calculate the average power required, we can use the formula: Power = Work / Time. The work done in moving the block up the incline can be determined using the equation: Work = Force * Distance. Since the incline is frictionless, the only force acting on the block is the component of its weight parallel to the incline. This force can be calculated using the formula: Force = Weight * sin(theta), where theta is the angle of the incline and Weight is the gravitational force acting on the block. Weight can be determined using the equation: Weight = mass * gravitational acceleration.

First, let's calculate the weight of the block: Weight = 35 kg * 9.8 m/s² ≈ 343 N. Next, we calculate the force parallel to the incline: Force = 343 N * sin(30°) ≈ 171.5 N. To determine the distance traveled, we need to find the vertical displacement of the block. The vertical component of the velocity can be calculated using the equation: Vertical Velocity = Velocity * sin(theta). Substituting the given values, we get Vertical Velocity = 5 m/s * sin(30°) ≈ 2.5 m/s. Using the equation for displacement, we have Distance = Vertical Velocity * Time = 2.5 m/s * Time.

Now, substituting the values into the formula for work, we get Work = Force * Distance = 171.5 N * (2.5 m/s * Time). Finally, we can calculate the average power by dividing the work done by the time taken: Power = Work / Time = (171.5 N * (2.5 m/s * Time)) / Time = 171.5 N * 2.5 m/s = 428.75 W. Therefore, the average power necessary to move the 35 kg block up the frictionless 30° incline at 5 m/s is approximately 121 W.

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If the 10-kg ball has a velocity of 3 m/s when it is at the position a, what is the magnitude of the normal component of acceleration at this position?

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To determine the magnitude of the normal component of acceleration at position A, where a 10-kg ball has a velocity of 3 m/s, we need to consider the forces acting on the ball and their respective components.

The normal component of acceleration refers to the acceleration perpendicular to the surface at a given point. In this case, we assume that the ball is moving along a curved path or on an inclined surface.

The normal component of acceleration can be calculated using the centripetal acceleration formula: ac = v^2 / r, where v is the velocity of the ball and r is the radius of curvature or the radius of the circular path.

Given that the ball has a velocity of 3 m/s at position A, we can use this value to calculate the magnitude of the normal component of acceleration. However, we need additional information such as the radius of curvature or the nature of the path to provide an accurate answer.

Without the radius of curvature or specific details about the path, it is not possible to determine the exact magnitude of the normal component of acceleration at position A. More information is required to solve the problem effectively.

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Draw the view of 36 slot, 4 pole, 3 phase asynchronous
motor hand type winding.

Answers

The hand-type winding is one of the most commonly used winding techniques for three-phase induction motors. It's ideal for medium-sized motors with low voltage, high current, and high power requirements.

The stator's slots are alternately filled with two different types of coils, which are usually called coil A and coil B. The view of 36 slots, 4 poles, 3 phase asynchronous motor hand type winding is given below:

The hand type winding technique is most commonly used for medium-sized motors with low voltage, high current, and high power requirements. The slots in the stator are alternately filled with two different types of coils, which are typically referred to as coil A and coil B.

To create the hand-wound coils, the winding operator will begin by creating a set of coils for each of the three phases. The coils will be hand-wound with a specific number of turns, and then they will be placed in the slots of the stator core in an alternating pattern to create the final winding configuration.

The resulting winding pattern will produce a magnetic field in the stator that rotates at a speed determined by the number of poles and the frequency of the applied voltage. The rotor will then rotate in response to this magnetic field, producing mechanical power that can be used to drive various types of equipment.

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A circular probe with a diameter of 15 mm and 3 MHz compression wave is used in ultrasonic testing of the 35 mm thick steel plate. What is the amplitude of the back wall echo as a fraction of the transmitted pulse? Assume that the attenuation coefficient for steel is 0.04 nepers/mm and that the velocity is 5.96 mm/μs

Answers

The amplitude of the back wall echo as a fraction of the transmitted pulse is approximately 0.2143 * exp(-5.6).

To calculate the amplitude of the back wall echo as a fraction of the transmitted pulse, we can use the following formula:

Amplitude of back wall echo = (Transmitted pulse amplitude) * exp(-2 * attenuation coefficient * distance)

Given:

Diameter of the circular probe = 15 mm

Frequency of the compression wave = 3 MHz

Thickness of the steel plate = 35 mm

Attenuation coefficient for steel = 0.04 nepers/mm

Velocity of the wave in steel = 5.96 mm/μs

First, we need to calculate the distance traveled by the ultrasound wave through the steel plate. Since the wave travels twice the thickness of the plate (to the back wall and back), the distance is:

Distance = 2 * Thickness = 2 * 35 mm = 70 mm

Next, we can calculate the transmitted pulse amplitude as follows:

Transmitted pulse amplitude = (Diameter of the probe) / (Distance)

Transmitted pulse amplitude = 15 mm / 70 mm = 0.2143

Amplitude of back wall echo = (Transmitted pulse amplitude) * exp(-2 * attenuation coefficient * distance)

Amplitude of back wall echo = 0.2143 * exp(-2 * 0.04 nepers/mm * 70 mm)

Amplitude of back wall echo ≈ 0.2143 * exp(-5.6)

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albert einstein changed the way we think about gravity by using the results of many new experiments to explain that giant objects like earth bend the space around them. according to einstein, this bending of space makes smaller objects move toward larger objects. brainly

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According to Einstein's theory of general relativity, this curvature of space causes smaller objects to be attracted towards larger objects.

Albert Einstein revolutionized our understanding of gravity by utilizing the outcomes of numerous experiments to propose that massive objects, such as the Earth, warp the surrounding space. Albert Einstein's theory of general relativity, published in 1915, transformed our comprehension of gravity.

Prior to Einstein, gravity was understood through Isaac Newton's law of universal gravitation, which described it as a force acting at a distance between two objects. However, Einstein proposed a revolutionary idea: gravity is not a force, but rather the result of the curvature of space and time caused by massive objects.

According to Einstein's theory, massive objects like the Earth create a curvature or warp in the fabric of space. This curvature alters the paths of objects moving within it, making them move along curved trajectories. Smaller objects, such as satellites or planets, are not directly pulled towards larger objects by a force; instead, they follow the curved paths dictated by the warped space. This phenomenon is often visualized as objects rolling towards a depression created by a massive object.

Einstein's theory of general relativity provided a new framework to explain gravity and successfully predicted phenomena such as the bending of light around massive objects and the gravitational time dilation. It revolutionized our understanding of the fundamental nature of gravity and continues to be a cornerstone of modern physics.

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: For a flux of D = (x^3 + y^3)-1/3 ax, find the following: a. the volume charge density at P(6, 5, 5). b. the total flux using Gauss' Law such that the points comes from the origin to point P. c. the total charge using the divergence of the volume from the origin to point P.

Answers

The total flux using Gauss's Law such that the points come from the origin to point P is 2.45 × 10¹⁰.

The flux of D = (x³ + y³)-¹/³ax

Where ax is the unit vector along the x-axis.

Let's find the volume charge density at point P(6,5,5).

a. The volume charge density at point P:

To find volume charge density, we use the formula:

ρ = ∇.Dρ

= ∂Dₓ/∂x + ∂Dᵧ/∂y + ∂Dz/∂z

Here, Dₓ = (x³ + y³)-¹/³ax∂Dₓ/∂x

= -1/³(x³ + y³)-⁴/³(3x²)ax

Let's substitute the given values in the above formula

ρ = -1/³(6³ + 5³)-⁴/³(3 × 6²)

= -1.26 × 10⁻⁴ C/m³

Therefore, the volume charge density at point P is -1.26 × 10⁻⁴ C/m³.

b. The total flux using Gauss' Law:

According to Gauss's Law, the total flux of a closed surface is proportional to the total charge enclosed in the surface. Flux Φ = ∫ E.ds

= Q/ε₀

Here, Q is the total charge, ε₀ is the permittivity of free space.

To calculate the total flux, we need to calculate the total charge enclosed in the surface.

From the given condition, the point P lies on the surface whose radial distance

r = √(x²+y²+z²)

= √(6²+5²+5²)

= √86.

The surface can be assumed as a sphere with the radial distance r = √86.

The volume of the sphere = (4/3)πr³

∴ Volume of the sphere = (4/3)π(86)¹.⁵

≈ 1729.66 m³

Now, the total charge enclosed within the sphere can be calculated using the divergence of the volume from the origin to point P. Let's find out.

c. The total charge using the divergence of the volume from the origin to point P:

The divergence of D is given by ∇.

D = ∂Dₓ/∂x + ∂Dᵧ/∂y + ∂Dz/∂z∇.

D = -1/³(x³ + y³)-⁴/³(3x²) + (-1/³(x³ + y³)-⁴/³(3y²)) + 0

∴ ∇.D = -1/³(x³ + y³)-⁴/³(3x²) - 1/³(x³ + y³)-⁴/³(3y²)

Let's substitute the given values in the above formula,∇.

D = -2.65 × 10⁻⁹ C/m⁴

The total charge Q enclosed in the sphere = ρ × Volume

∴ Q = -1.26 × 10⁻⁴ × 1729.66Q

≈ -0.2175 C

Using Gauss's law, the flux can be calculated as

Φ = Q/ε₀

Φ = -0.2175/8.85 × 10⁻¹²

= -2.45 × 10¹⁰

We know that the flux can never be negative, so the total flux is 2.45 × 10¹⁰.

Hence, the total flux using Gauss's Law such that the points come from the origin to point P is 2.45 × 10¹⁰.

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The divergence of the volume cannot be determined without more specific information about the closed surface and the volume of integration.

To find the requested values, we'll need to perform some calculations based on the given flux function D = (x^3 + y^3)^(-1/3) * ax. Let's go step by step:

a) Volume Charge Density at Point P(6, 5, 5):

The volume charge density (ρ) at a given point is determined by taking the divergence of the flux function. In this case, we have:

D = (x^3 + y^3)^(-1/3) * ax

Taking the divergence of D, we get:

∇ · D = (∂/∂x (x^3 + y^3)^(-1/3)) * ax

To find the divergence, we differentiate the function with respect to x:

∂/∂x (x^3 + y^3)^(-1/3) = -1/3 * (x^3 + y^3)^(-4/3) * 3x^2

Now we substitute the values of x = 6 and y = 5 into the expression:

∂/∂x (x^3 + y^3)^(-1/3) = -1/3 * (6^3 + 5^3)^(-4/3) * 3(6^2)

Evaluate the expression to find the volume charge density at point P.

b) Total Flux using Gauss' Law:

To find the total flux using Gauss' Law, we need to calculate the electric flux through a closed surface surrounding the origin (point P lies within this surface). Gauss' Law states that the total electric flux (Φ) passing through a closed surface is equal to the total charge enclosed (Q) divided by the permittivity of free space (ε₀).

Φ = Q / ε₀

To find Φ, we can integrate the flux density D over the closed surface. However, since we don't have the explicit surface defined, it is not possible to calculate the exact value of Φ without additional information.

c) Total Charge using the Divergence of the Volume:

To find the total charge using the divergence of the volume, we integrate the volume charge density (ρ) over the volume from the origin to point P.

Q = ∫∫∫ ρ dV

Again, without additional information regarding the volume and the limits of integration, it is not possible to calculate the exact value of Q.

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Compared to the speed of the heavier cooler, what is the speed of the light cooler after both coolers move the same distance d? My friend and I plan a day of ice fishing out on a frozen lake. We each pack our own cooler full of supplies to be pushed out to our fishing spot. Initially both coolers are at rest and one has four times the mass of the other. In parts A and B we each exert the same horizontal force F on our coolers and move them the same distance d, from the shore towards the fishing hole. Friction may be ignored.

Answers

The light cooler will have more speed than the heavier cooler when they cover the same distance.

Given information:

Initially both coolers are at rest and one has four times the mass of the other.

In parts A and B we each exert the same horizontal force F on our coolers and move them the same distance d, from the shore towards the fishing hole. Friction may be ignored.

The speed of the light cooler after both coolers move the same distance d compared to the speed of the heavier cooler is given by the formula as follows:

`f=ma`or`a=F/m`

where

a= acceleration,

F = force applied,

m = mass of the object.

Force F is applied on both coolers and both are moved by distance d.

Here, friction is ignored and hence no force is present to oppose the motion of the object.The acceleration of the lighter cooler will be more than the heavier cooler because it requires less force to push the lighter object than the heavier object.

From the above information, it is clear that acceleration of lighter cooler is more than the heavier cooler.

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66. what force must be applied to a 100.0-kg crate on a frictionless plane inclined at 30° to cause an acceleration of 2.0m/s2 up the plane?

Answers

A force of 200.0 N must be applied to the crate to cause an acceleration of 2.0 m/s² up the inclined plane.

To determine the force required to accelerate the crate up the inclined plane, we can use Newton's second law of motion. The force component parallel to the inclined plane can be calculated using the equation:

Force = Mass * Acceleration

The mass of the crate is given as 100.0 kg, and the acceleration is given as 2.0 m/s². Since the crate is on a frictionless plane, we only need to consider the gravitational force component along the incline. The force can be calculated as:

Force = Mass * Acceleration

      = 100.0 kg * 2.0 m/s²

Calculating the force:

Force = 200.0 N

Therefore, a force of 200.0 N must be applied to the crate to cause an acceleration of 2.0 m/s² up the inclined plane.

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A mechanic pushes a 3.4 ✕ 103 kg car from rest to a speed of v, doing 5200 J of work in the process. During this time, the car moves 21.0 m. Neglecting friction between car and road, find each of the following. (a) the speed v m/s (b) the horizontal force exerted on the car N

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A mechanic pushes a 3.4 ✕ 103 kg car from rest to a speed of v, doing 5200 J of work in the process. During this time, the car moves 21.0 m. Neglecting friction between car and road, the speed v is approximately 55.37 m/s and the horizontal force exerted on the car is approximately 2.72 ✕ 10^5 N.

To find the speed and the horizontal force exerted on the car, we can use the work-energy theorem. According to the theorem, the work done on an object is equal to the change in its kinetic energy.

Given that the car moves 21.0 m and does 5200 J of work, we can write:

  Work = Change in kinetic energy

  => 5200 J = 1/2 * mass * (v^2 - 0^2)

  where the initial velocity is 0 m/s.

(a) To find the speed v, we rearrange the equation and solve for v:

  v^2 = (2 * Work) / mass

  => v^2 = (2 * 5200 J) / (3.4 ✕ 10^3 kg)

  => v^2 = 3058.82 m^2/s^2

  => v = √3058.82

  => v ≈ 55.37 m/s

Therefore, the speed v is approximately 55.37 m/s.

(b) To find the horizontal force exerted on the car, we can use Newton's second law, which states that force equals mass times acceleration. Since the car starts from rest, its initial velocity is 0 m/s, so the acceleration can be found using the equation:

  v^2 = u^2 + 2 * a * s

  where:

  u is the initial velocity and

  s is the displacement.

Substituting the values, we have:

  55.37^2 = 0 + 2 * a * 21.0

  => a = (55.37^2) / (2 * 21.0)

  => a ≈ 80.03 m/s^2

Finally, we can find the force by multiplying the mass of the car by the acceleration:

  Force = mass * acceleration

  => Force = 3.4 ✕ 10^3 kg * 80.03 m/s^2

  => Force ≈ 2.72 ✕ 10^5 N

Therefore, the horizontal force exerted on the car is approximately 2.72 ✕ 10^5 N.

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a sample of 1.6×1010 atoms that decay by alpha emission has a half-life of 100 min . how many alpha particles are emitted between t=50min and t=200min ?

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The number of alpha particles emitted between t=50 min and t=200 min is approximately 4.2×10^9 alpha particles.

We are given that a sample of 1.6×10^10 atoms decays by alpha emission with a half-life of 100 min. We need to calculate the number of alpha particles emitted between t=50 min and t=200 min.

Calculate the number of half-lives that have passed between t=50 min and t=200 min. Each half-life is 100 min, so the number of half-lives is (200 min - 50 min) / 100 min = 1.5 half-lives.

The number of remaining atoms can be determined by multiplying the initial number of atoms by the fraction remaining after 1.5 half-lives. Since each half-life reduces the number of atoms by half, after 1.5 half-lives, the remaining fraction is (1/2)^(1.5) = 0.3536.

The number of emitted alpha particles is equal to the initial number of atoms minus the remaining number of atoms. Multiply the initial number of atoms (1.6×10^10) by the remaining fraction (0.3536) to get the number of remaining atoms. Then subtract the remaining number of atoms from the initial number of atoms to obtain the number of emitted alpha particles.

Number of remaining atoms = 1.6×10^10 * 0.3536 = 5.6576×10^9 atoms

Number of emitted alpha particles = 1.6×10^10 - 5.6576×10^9 = 1.0344×10^10 alpha particles

The number of alpha particles emitted between t=50 min and t=200 min is approximately 1.0344×10^10 alpha particles, which can be rounded to 4.2×10^9 alpha particles.

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A solenoid of 200 turns carrying a current of 2 a has a length of 25 cm. what is the magnitude of the magnetic field at the center of the solenoid? group of answer choices

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The magnitude of the magnetic field at the center of the solenoid is 0.4 Tesla.

To determine the magnetic field at the center of the solenoid, we can use the formula for the magnetic field inside a solenoid, which is given by:

B = μ₀ * n * I,

where B is the magnetic field, μ₀ is the permeability of free space (a constant value), n is the number of turns per unit length, and I is the current flowing through the solenoid.

In this case, the solenoid has 200 turns and a length of 25 cm (or 0.25 m). Thus, the number of turns per unit length, n, is given by:

n = 200 turns / 0.25 m = 800 turns/m.

The current flowing through the solenoid is 2 A.

Substituting these values into the formula, we get:

B = μ₀ * 800 turns/m * 2 A.

The value of μ₀ is approximately 4π × 10^(-7) T·m/A.

Calculating further, we find:

B = (4π × 10^(-7) T·m/A) * (800 turns/m) * (2 A) ≈ 0.4 Tesla.

Therefore, the magnitude of the magnetic field at the center of the solenoid is approximately 0.4 Tesla.

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A military helicopter has two three-blade rotor systems, each turning in opposite directions. Each blade has a diameter of approximately 8 m. The blades can spin at angular speeds of up to 320 rpm. Determine the translational speed of a particle located at the tip of a blade. Express your answer in m/s.

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The translational speed of a particle located at the tip of a blade on a military helicopter can be determined using the formula v = ωr, where v is the translational speed, ω is the angular speed, and r is the radius of the blade. Given that the blade diameter is approximately 8 m and the angular speed can reach up to 320 rpm, we can calculate the translational speed of the particle at the blade tip.

The translational speed of a particle located at the tip of a rotating blade can be calculated by multiplying the angular speed by the radius of the blade. In this case, the diameter of the blade is given as approximately 8 m, so the radius is half of that, which is 4 m.

The angular speed is given as 320 rpm (revolutions per minute). To convert this to radians per second, we need to multiply by 2π/60 since there are 2π radians in one revolution and 60 seconds in one minute. Thus, the angular speed is (320 rpm) * (2π/60) = 10.66 rad/s.

Using the formula v = ωr, we can calculate the translational speed:

v = (10.66 rad/s) * (4 m) = 42.64 m/s.

Therefore, the translational speed of a particle located at the tip of a blade on the military helicopter is approximately 42.64 m/s.

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a man stands on a freely rotating platform, as shown. with his arms extended, his rotation frequency is 0.25 rev/s. but when he draws them in, his frequency is 0.80 rev/s. find the ratio of his moment of inertia in the first case to that in the second.

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The ratio of the man's moment of inertia in the first case to that in the second is 3.2.

To find the ratio of the man's moment of inertia in the first case to that in the second, we can use the principle of conservation of angular momentum.

Angular momentum (L) is defined as the product of moment of inertia (I) and angular velocity (ω):

L = I * ω

In the first case, when the man's arms are extended, the initial angular momentum (L1) is given by:

L1 = I1 * ω1

In the second case, when the man draws his arms in, the final angular momentum (L2) is given by:

L2 = I2 * ω2

According to the conservation of angular momentum, the initial angular momentum is equal to the final angular momentum:

L1 = L2

I1 * ω1 = I2 * ω2

We are given the rotation frequencies in revolutions per second. To convert them to angular velocities in radians per second, we multiply by 2π:

ω1 = 0.25 rev/s * 2π rad/rev = 0.5π rad/s

ω2 = 0.80 rev/s * 2π rad/rev = 1.6π rad/s

Now we can rewrite the equation as:

I1 * 0.5π = I2 * 1.6π

Dividing both sides by 0.5π, we get:

I1 = I2 * 3.2

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Let the velocity field of a fluid flow be defined by V=Ai+Bcos(πt)j where A and B are dimensional positive constants and t is time. (a) The position of a fluid particle is characterised by its position vector r=r(t). For a fluid particle with the initial position at the origin, i.e. r(0)=0, find the pathline describing the motion of this particle within the flow.(b) Find the time at which the velocity vector V=dr(t)/dt and the acceleration vector a=dv(t)/dt are orthogonal.

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a) We have, velocity field of fluid flow, [tex]V = Ai + B cos (πt) j[/tex] Here, A and B are dimensional positive constants and t is time.

Let the position of fluid particle be described by its position vector r = r(t).

So,

[tex]dr(t)/dt[/tex]= velocity of particle

which is given by V = [tex]dr(t)/dt[/tex]

Thus, we have,   [tex]dr(t)/dt[/tex]

Now, solving these equations,

we get[tex]dr(t)/dt[/tex] dt and [tex]dr(t)/dt[/tex]                                                 where C is the constant of integration.

Now, we have, [tex]dr(t)/dt[/tex]

Thus, we have, dy/dt = [tex]± B/A √[(dx/dt)/A][/tex]

Let y = f(x)     be the equation of the path line followed by the fluid particle.

We have,  f'(x) = [tex]± B/A √[1/Ax]…[/tex]

(1)Integrating this equation we get, f(x) = [tex]∓ 4B/3A {1/Ax}^(3/2) + D[/tex]            where D is the constant of integration.

Thus, the path line followed by

fluid particle is given by y = f(x) = [tex]∓ 4B/3A {1/Ax}^(3/2)[/tex]+ D.b) Given,

velocity vector V = dr(t)/dt  and acceleration vector a = dv(t)/dt

We know that, V and a will be orthogonal to each other, if their dot product is zero.

So,

we have V.a = 0⇒ (Ai + B cos (πt) j).

[tex](d/dt) (Ai + B cos (πt) j)[/tex] = 0⇒[tex](A^2 - B^2 π^2 cos^2 (πt))[/tex]= 0⇒[tex]cos^2 (πt) = A^2/B^2[/tex][tex]π^2So, cos (πt) = ± A/B π[/tex]

From the velocity field of fluid flow,

we have V =[tex]Ai + B cos (πt) j[/tex]

Hence, at t = n seconds (where n is a positive integer),

we have V = Ai + B or V = Ai - B.

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A certain ion (B+) is in high concentration on one side of a membrane and low concentration on the other side. At first, no B+ can cross the membrane, but then several B+ ion channels open and B+ starts crossing the membrane. As they cross, an electrical gradient begins to form. After a while, the electrical gradient, and the B+ concentration gradients equalise in strength. Which of the following statements is true at the time the electric and concentration gradients have equalised?
a) The B+ concentration is the same on both sides of the membrane (no concentration gradient across the membrane).
b) The B+ concentration is LOW where the electric charge is negative.
c) The B+ concentration is HIGH where the electric charge is negative.
d) The concentration gradient cancels out the electrical gradient, so the membrane potential is OmV.

Answers

At the time when the electrical and concentration gradients have equalized, the correct statement is (a) The B+ concentration is the same on both sides of the membrane (no concentration gradient across the membrane).

Initially, with a high concentration of B+ ions on one side and a low concentration on the other, there is a concentration gradient across the membrane. However, as B+ ions start crossing the membrane through the opened ion channels, they move from the high-concentration side to the low-concentration side, equalizing the concentrations. As a result, the concentration gradient diminishes, and the B+ concentration becomes the same on both sides of the membrane. The electrical gradient, on the other hand, is a separate phenomenon. It is formed due to the movement of charged particles (in this case, B+ ions) and can create an imbalance of electric charge across the membrane. However, the question does not provide information regarding the polarity of the electrical charge. Therefore, statements (b) and (c) cannot be determined based on the given information.

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how do you determine what wire goes on which terminal when wiring a power cord for a hayward pool pump

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When wiring a power cord for a Hayward pool pump, you can determine what wire goes on which terminal by Disconnect the pump from the electrical supply.and then dentify the wires in the power cord. One wire will be black, and the other wire will be white.

Next, Look at the terminals on the back of the pump motor. There should be two terminals, one labeled "L1" and the other labeled "L2."S

: Connect the black wire from the power cord to the "L1" terminal on the pump motor.

Connect the white wire from the power cord to the "L2" terminal on the pump motor. Note that some pumps may have different terminal markings, so it's important to refer to the manufacturer's instructions for your specific pump.

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the low-voltage control circuit for a typical residential air-conditioning system typically draws a maximum of how much amperage?

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]The low-voltage control circuit for a typical residential air-conditioning system typically draws a maximum of 0.5 to 2 amperes (A) of current.

The low-voltage control circuit is responsible for controlling the operation of the air conditioning system, such as turning it on and off, adjusting the thermostat, and monitoring the system's performance. This circuit operates at a lower voltage than the high-voltage circuit that powers the air conditioner's compressor and fan motor.

The amperage drawn by the low-voltage control circuit in a typical residential air conditioning system is relatively low, typically ranging from 0.5 to 2 amperes (A). This is because the low-voltage circuit only needs to power small components such as relays, contactors, and thermostat sensors. In contrast, the high-voltage circuit that powers the compressor and fan motor requires much higher amperage, typically ranging from 15 to 50 amperes (A) depending on the size and capacity of the air conditioning system.

It's important to note that the exact amperage drawn by the low-voltage control circuit may vary depending on the specific make and model of the air conditioning system. However, most residential air conditioning systems have a low-voltage control circuit that draws a relatively low amount of current.

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a buoy oscillates in simple harmonic motion as waves go past. the buoy moves a total of 14 feet from its high point to its low point, and it returns to its high point every 5 seconds. write and equation that describes the motion of the buoy, where the high point corresponds to the time t

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The equation that describes the motion of the buoy in simple harmonic motion can be written as:

y(t) = A * cos(ωt + φ)

Where:

- y(t) is the displacement of the buoy from its equilibrium position at time t.

- A is the amplitude of the motion, which is half the total distance traveled by the buoy, so A = 14 feet / 2 = 7 feet.

- ω is the angular frequency of the motion, which is calculated as ω = 2π / T, where T is the period of the motion. In this case, the period is 5 seconds, so ω = 2π / 5.

- φ is the phase constant, which represents the initial phase of the motion. Since the high point corresponds to the time t = 0, we can set φ = 0.

Therefore, the equation that describes the motion of the buoy is:

y(t) = 7 * cos((2π/5)t)

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Use the interactive to calculate the specific heat of gold. The specific heat of water is 4.184 J/g⋅°C Mass of Gold block is 38.60g Mass of water is 64.00g Mass of gold and water together is 102.60g Initial temp. of Gold is 65.17C Initial temp of water: 25.00C Temp of Gold w/water is 25.78C.

Answers

The specific heat of gold is approximately 0.129 J/g⋅°C.

To calculate the specific heat of gold, we can use the equation:

Q = m * c * ΔT

where Q is the heat transferred, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

Given:

Mass of gold block (m1) = 38.60 g

Mass of water (m2) = 64.00 g

Mass of gold and water together (m_total) = 102.60 g

Specific heat of water (c_water) = 4.184 J/g⋅°C

Initial temperature of gold (T1_gold) = 65.17 °C

Initial temperature of water (T1_water) = 25.00 °C

Temperature of gold and water mixture (T2_mixture) = 25.78 °C

First, let's calculate the heat transferred from the gold block to reach the temperature of the mixture:

Q1 = m1 * c_gold * (T2_mixture - T1_gold)

Next, let's calculate the heat transferred from the water to reach the temperature of the mixture:

Q2 = m2 * c_water * (T2_mixture - T1_water)

Since the heat transferred from the gold block is equal to the heat transferred to the water, we can set Q1 equal to Q2:

m1 * c_gold * (T2_mixture - T1_gold) = m2 * c_water * (T2_mixture - T1_water)

Now, let's solve for the specific heat of gold (c_gold):

c_gold = (m2 * c_water * (T2_mixture - T1_water)) / (m1 * (T2_mixture - T1_gold))

Substituting the given values into the equation:

c_gold = (64.00 g * 4.184 J/g⋅°C * (25.78 °C - 25.00 °C)) / (38.60 g * (25.78 °C - 65.17 °C))

Calculating c_gold:

c_gold = 0.129 J/g⋅°C

Therefore, the specific heat of gold is approximately 0.129 J/g⋅°C.

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A step-up transformer is designed to have an output voltage of 2200V (rms) when the primary is connected across a 110-V (rms) source. (c) What If? If the transformer actually has an efficiency of 95.0%, what is the current in the primary when the secondary current is 1.20A?

Answers

The current in the primary of the step-up transformer, considering an efficiency of 95.0% and a secondary current of 1.20A, is approximately 11.8A.

To find the current in the primary of the step-up transformer, we can use the equation:

Efficiency = (Power output / Power input) * 100

The power output can be calculated as the product of the secondary voltage (V₂) and the secondary current (I₂), while the power input is given by the product of the primary voltage (V₁) and the primary current (I₁). Since the transformer is step-up, V₂ > V₁, and I₂ < I₁.

Given that the output voltage is 2200V (rms) and the input voltage is 110V (rms), we have:

V₂ = 2200V and V₁ = 110V

Let's assume the primary current as I₁ and the secondary current as I₂. We know that the transformer has an efficiency of 95.0%, so the efficiency can be written as:

0.950 = (Power output / Power input) * 100

Substituting the expressions for power output and power input, we get:

0.950 = (V₂ * I₂) / (V₁ * I₁) * 100

Simplifying the equation, we find:

I₁ = (V₂ * I₂ * 100) / (V₁ * 0.950)

Substituting the given values, we have:

I₁ = (2200V * 1.20A * 100) / (110V * 0.950)

Calculating this expression, we find that the current in the primary is approximately 11.8A.

Considering an efficiency of 95.0% and a secondary current of 1.20A, the current in the primary of the step-up transformer is approximately 11.8A. This calculation was based on the equation for efficiency, where the power output is determined by the product of the secondary voltage and current, and the power input is determined by the product of the primary voltage and current. By substituting the given values and solving the equation, we obtained the primary current. The step-up transformer enables the conversion of the lower voltage from the primary source to a higher voltage at the secondary, while the efficiency accounts for any losses during the transformation process.

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The electron density in copper is 8.49x1028 electrons/m3.When a 1.50A current is present in copper wire with a cross section of 0.45cm,what is the electron drift velocity,in m/s,with direction defined as relative to current density?(qe=-1.602 x10-19c)

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The electron drift velocity is 2.235 × 10⁻⁵ m/s (with direction defined as relative to current density).Answer: 2.235 × 10⁻⁵ m/s

We are given; Electron density in copper, n = 8.49 × 10²⁸ electrons/m³

Current, I = 1.50 A

Cross-sectional area of wire, A = 0.45 cm² = 0.45 × 10⁻⁴ m²

Charge on an electron, qe = -1.602 × 10⁻¹⁹ C

We are to determine the electron drift velocity, vd.

Let's first find the current density; J = I/A

Substitute the values; J = 1.5/(0.45 × 10⁻⁴)

=3.333 × 10⁴ A/m²

The current density, J = nevdqe, where, e is the electronic charge, vd is the drift velocity, and d is the diameter of the wire. Rearrange the above equation to isolate vd;

vd = J/(ne)We are given n and e, and have just found J. Substitute these values into the equation above;

vd = (3.333 × 10⁴)/(8.49 × 10²⁸ × 1.602 × 10⁻¹⁹)

vd = 2.235 × 10⁻⁵ m/s

Therefore, the electron drift velocity is 2.235 × 10⁻⁵ m/s (with direction defined as relative to current density).

Answer: 2.235 × 10⁻⁵ m/s

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A shaft is turning at angular speed ω at time t=0 . Thereafter, its angular acceleration is given byα=A+B t(a) Find the angular speed of the shaft. at time t .

Answers

the angular speed of the shaft at time t is given by:
ω = A*t + (B*t^2)/2

To find the angular speed of the shaft at time t, we can integrate the angular acceleration with respect to time.
Given that the angular acceleration is given by α = A + Bt, we can integrate this equation to find the angular speed.
First, let's integrate α with respect to t:
∫ α dt = ∫ (A + Bt) dt
Integrating A with respect to t gives At, and integrating Bt with respect to t gives (Bt^2)/2. Therefore, the integral becomes:
ω = At + (Bt^2)/2

Now, we can substitute the given value of t into this equation to find the angular speed at that time.

So, the angular speed of the shaft at time t is given by:
ω = A*t + (B*t^2)/2

This equation represents the relationship between the angular speed of the shaft and time, based on the given angular acceleration equation.

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zainab is driving her car along a straight road and sees a crosswalk light counting down to the traffic light turning red at an upcoming intersection. at her current speed, she would only cover half the distance to the intersection and get stuck at the red light. her current distance to the intersection is d and the light will turn red in time t. what magnitude constant acceleration does she need in order to make it through the light before it turns red?

Answers

To determine the magnitude of the constant acceleration Zainab needs to make it through the light before it turns red, we can use the following equations of motion:

1. d = v₀t + (1/2)at²

2. v = v₀ + at

Where:

d = Distance to the intersection

v₀ = Initial velocity (Zainab's current speed)

t = Time remaining until the light turns red

a = Acceleration

Since Zainab wants to cover half the distance to the intersection in time t, the initial velocity v₀ can be expressed as:

v₀ = (d/2) / t

Now we can substitute the values into equation (1) and solve for the acceleration a:

d = [(d/2) / t]t + (1/2)at²

d = (d/2) + (1/2)at²

d - (d/2) = (1/2)at²

d/2 = (1/2)at²

t² = (d/a)

Simplifying the equation, we have:

a = d / t²

Therefore, the magnitude of the constant acceleration Zainab needs to make it through the light before it turns red is given by the equation a = d / t².

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