The secondary line-to-line voltage of a three-phase Y-Y-connected transformer bank is 115 VAC. What would be the secondary line to neutral voltage?


A. 199.2 V

B. 66.4 V

C. 84.2 V

D. 50 V

Answer:

Coefficient of friction is

Ū = 0.31

Explanation:

T2 = T1* e^(ūơ)

Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,

ū = coefficient of friction

Ơ = 2pai * N

Where N = number of turns = 2

Ơ = 4pai

7500 = 150e^(ū*4pai)

50 = e^(ū *4pai)

lin 50 = 4pai * ū

Ū = 3.91/4pai

Ū = 0.31

Answer:

x -component of dog's velocity = 8.265m/s

y- component of dog's velocity = 1.389m/s

Explanation:

The detailed and step by step calculation is as shown in the attached file.

Answer and Explanation

It was initially specified that the receptacle outlet to be located within 50 ft of the electrical service equipment.

As well, instead of the receptacle being required within 50 ft., it is now required within 25 ft. of the service. This was to accommodate the typical 25 ft. cord used by many service electricians.

The rules apply to indoor service locations other than one-and two-family dwellings. For these locations, at least one 125-volt, single-phase, 15- or 20-ampere receptacle outlet must be installed in an accessible location within 25 ft. of the indoor electrical service equipment. The receptacle must be within the same room or area as the actual service equipment.

Having a maintenance receptacle near the electrical service allows for testing, servicing and connection of portable electrical data acquisition equipment for analyzing the electrical system.

The phasor currents for the load are 0.573 per unit. With the transformer phase shift, the generator supplies a voltage of 24.3 KV and 54.6 MW. Ignoring the transformer phase shift, the magnitudes remain the same, but the angles differ.

The complex power supplied to the load is P = 240 MVA * 0.9 = 216 MW and reactive power Q = 240 MVA * sqrt (1-0.9^2) = 106.3 MVAR. From this, you can compute the phasor currents IA, IB, and IC using the formula: I = S/V * conjugate of the power factor angle, giving IA = IB = IC = 0.573 per unit.

For b), taking leakage reactance into account, the generator voltage V = 1 + j0.11 * I, giving a voltage of 1.063 per unit. Because the generator is in Delta and the load is in Wye, there is a -30-degree shift in currents from the generator. So, IA, IB, and IC are 0.573 angle -30 degrees per unit.

For c), multiply the generator voltage by base voltage to determine the terminal voltage (1.063*23kV = 24.3 kV). The total real power supplied by the generator PG = VI*cos(theta) = 0.573*1.063*0.9 = 0.546 per unit, or 54.6 MW.

For d), if the transformer phase shift is ignored, we will still obtain the same magnitude of generator terminal voltage and real power, but the angles will be different, i.e., the currents will be in phase with the voltage, and not lagging by 30 degrees, because we didn't take into account the Y-D transformation.

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Answer

given,

mass of the rock, m = 30 Kg

Length of the string, r = 0.25 m

angular speed of the rock, ω = 12 rad/s

Tension in the string at the top of the vertical circle = ?

At the top of the string the tension is

[tex]T_{top}=\dfrac{mv^2}{r} - m g[/tex]

we know     v = r ω

[tex]T_{top}= mr\omega^2 - m g[/tex]

[tex]T_{top}= 30\times 0.25\times 12^2-30\times 9.8[/tex]

[tex] T_{top} = 786 N[/tex]

if the given mass is 0.3 Kg then tension in the rope

[tex]T_{top}= 0.30\times 0.25\times 12^2-0.30\times 9.8[/tex]

[tex]T_{top} = 7.86 N[/tex]

If the mass is 0.3 then the correct answer is option A.

Answer: A CD rotate at a varying angular speed

A CD rotates at varying angular speed despite having a constant linear speed. This variation in rotation speed is necessary for the CD to operate efficiently.

Unlike a phonograph record, a CD rotates at varying angular speed even though it scans information at a constant linear speed due to the layout of data on the disc. The rotation speed decreases as the laser moves from the inside to the outside of the CD, maintaining a constant linear speed and adjusting the angular speed accordingly. This change in angular speed is essential for the CD to function effectively.

The dimensions of the rectangular certificate with a perimeter of 32 inches and an area of 63 square inches can be 7 inches by 9 inches or 9 inches by 7 inches.

The subject of this question is Mathematics, specifically an application of algebra to solve for the dimensions of a rectangle when given its perimeter and area. Let’s denote length as L and width as W. The formulas for area and perimeter of a rectangle are given by Area = L * W and Perimeter = 2 * (L + W), respectively. Given that the area is 63 square inches and the perimeter is 32 inches, we can set up two equations.

From the perimeter, we have 2L + 2W = 32, simplifying gives us L = 16 - W.

We can then substitute this into the area equation, so (16 - W) * W = 63.

This simplifies and solved that gives us W = 7 or W = 9.

To find L, we substitute W = 7 / W = 9 into L = 16 - W. We get the pairs (L, W) as (9, 7) or (7, 9).

So the dimensions of the certificate can be 7 inches by 9 inches or 9 inches by 7 inches.

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Answer

a) -2842 J

b) 2842 J

Explanation:

The escalator is moving the man of mass 58 kg till the height of 5 meters

So here;

m=58 kg

distance = height = h= 5 m

gravitational acceleration = g= 9.8 m/sec2

a) We know work is scalar product of force and displacement

i-e Work = F.S= FS cosθ, where θ is the angle between force vector and displacement vector

When escalator is moving up - the gravitational force is acting downward i- opposite to displcacement i-e angle between force and displacement is 180 degrees.

we have data

m= 58 kg

F=mg = 58×9.8=568.4 N

S=5 m

Work = F.S=FS cos 180°=568.4×5×(-1)= -2842 J

So the work done by gravitational force  is -2842 J and -ve sign here indicates that  distance is traveled in opposite direction of force.

b) When escalator is moving up the,   force is exerted by the escalator equal to gravitational force and the displacement is in the same direction so the angle between force and displacement is 0 degrees

Work = F.S=FS cos (0)=568.4×5×1=2842 J

So work done by the gravitational force then will be 2842 J

So in both cases work is same but in opposite directions.

Answer:From whom should a departing VFR aircraft request radar traffic information during ground operations? ... Answer: Sequencing to the primary Class C airport, traffic advisories, conflict resolution, and safety alerts.

Explanation:

Final answer:

During ground operations, a departing VFR aircraft should request radar traffic information from ground control. Once airborne, the responsibility switches to tower control for traffic updates.

Explanation:

A departing VFR (Visual Flight Rules) aircraft should request radar traffic information from the ground control (also known as ground movement control) before taxiing. Ground control is responsible for the safety on the taxiways and runways, except the active runway. Once airborne or when holding short of the runway, the aircraft switches to tower frequency, at which point the aircraft may receive radar traffic updates relevant to their departure.

Technology utilizing a network of radio frequency waveguides is part of managing air traffic, which is crucial for maintaining a safe separation between aircraft. Furthermore, radar technology can provide directional information, and as noted in research such as Hasselmann, 1991, traffic information can be directly calculated from radar data even during the ground operations of VFR aircraft.

Answer:

[tex]s=5.79\ km[/tex]

[tex]\theta=47^{\circ}[/tex] east of south

Explanation:

Given:

Now refer the schematic for visualization of situation:

[tex]y=d'.\sin47^{\circ}-d[/tex]

[tex]y=8.4\times \sin47-5.3[/tex] ...............(1)

[tex]x=d'.\cos47^{\circ}[/tex]

[tex]x=8.4\times \cos47^{\circ}[/tex] .................(2)

Now the direction of the desired position with respect to south:

[tex]\tan\theta=\frac{y}{x}[/tex]

[tex]\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}[/tex]

[tex]\theta=47^{\circ}[/tex] east of south

Now the distance from the current position to the desired position:

[tex]s=\sqrt{x^2+y^2}[/tex]

[tex]s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}[/tex]

[tex]s=5.79\ km[/tex]

Answer: False

Explanation:

It is dependent on the intensity of the radiation i.e the photo electron should increase with the light amplitude.

Answer with Explanation:

Let r be the resistance of short piece of copper wire.

Resistance of copper wire=R=[tex]13\Omega[/tex]

Resistance is directly proportional to length.

If a wire has greater resistance then,the wire will be greater in length.

Therefore,resistance of long piece of wire=7r

Total resistance of copper  wire=Sum of resistance of two piece of wires

[tex]r+7r=13[/tex]

[tex]8r=13[/tex]

[tex]r=\frac{13}{8}[/tex]ohm

Resistance of long piece of wire=[tex]7\times\frac{13}{8}=\frac{91}{8}\Omega[/tex]

Resistance of short piece of wire =[tex]\frac{13}{8}\Omega[/tex]

Resistivity of wire and cross section area of wire remains same .

Let L be the total  length  of wire and L' be the length of short  piece of wire.

We know that

[tex]R=\frac{\rho L}{A}[/tex]=[tex]\frac{\rho}{A}L=KL[/tex]

[tex]\frac{R}{L}=K[/tex]

Where K=[tex]\frac{\rho}{A}[/tex]=Constant

Using the formula

[tex]\frac{13}{L}=\frac{\frac{13}{8}}{L'}[/tex]

[tex]\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}[/tex]

[tex]L'=\frac{L}{8}[/tex]

Length of short piece of wire=L'=[tex]\frac{L}{8}[/tex]

Length of long piece of  wire=[tex]L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L[/tex]

% of length of short piece of   wire=[tex]\frac{\frac{L}{8}}{L}\times 100=12.5%[/tex]%

The resistance of the short piece=[tex]\frac{13}{8}\Omega[/tex]

The resistance of the long piece=[tex]\frac{91}{8}\Omega[/tex]

The length of short piece wire is [tex]12.5[/tex] % of total wire length.

The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex]

The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]

Let us consider that resistance of short piece wire is r.

So that, resistance of long piece wire is [tex]7r[/tex].

Total resistance of wire is [tex]13[/tex] ohms.

              [tex]7r+r=13\\\\8r=13\\\\r=\frac{13}{8}Ohm[/tex]

Let us consider that total length of wire is 13L.

So that, length of short piece wire [tex]=\frac{13}{8} L[/tex]

length of short piece wire is,

                 [tex]=\frac{\frac{13}{8}L }{13L}*100 =\frac{1}{8} *100=12.5[/tex]

The length of short piece wire is 12.5 % of total wire length.

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Answer and Explanation

The concept of parallax and relative motion is responsible for this.

That is, the photographer's motion prompts her to see parallax for the man and the trees, because their positions appear to keep shifting even though they are not really moving.

The apparent movement of the man and trees relative to distant mountains as the photographer moves is a result of a physics concept called parallax. Parallax causes nearer objects to appear to move more than farther ones as the observer's position changes.

The phenomenon you're describing occurred as a result of a concept in physics known as parallax. Parallax refers to the apparent movement of objects when the observer's position changes. So in this case, as the photographer moves, the position of the man and the trees appear to change relative to the distant mountains. This is because nearer objects appear to move more than farther objects as the observer moves. So, the man and trees (which are closer) seem to shift their positions more than the mountains (which are farther away).

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Answer:

b) twice the energy of each photon of the red light.

Explanation:

[tex]\lambda[/tex] = Wavelength

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

Energy of a photon is given by

[tex]E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}[/tex]

Let [tex]\lambda_1[/tex] = 700 nm

[tex]\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}[/tex]

For red light

[tex]E_1=\dfrac{hc}{\lambda_1}[/tex]

For UV light

[tex]E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}[/tex]

Dividing the equations

[tex]\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1[/tex]

Hence, the answer is  b) twice the energy of each photon of the red light.

Answer:

b) twice the energy of each photon of the red light.

Explanation:

Given:

wavelength of red light, [tex]\lambda_r=7\times 10^{-7}\ m[/tex]

wavelength of ultraviolet light, [tex]\lambda_{uv}=3.5\times 10^{-7}\ m[/tex]

We know that the energy of a photon is given as:

[tex]E=h.\nu[/tex] ..............(1)

where:

h = plank's constant [tex]=6.626\times 10^{-34}\ J.s[/tex]

[tex]\nu=[/tex]  frequency of the wave

we have the relation:

[tex]\nu=\frac{c}{\lambda}[/tex] ...................(2)

From (1) and (2) we have:

[tex]E=\frac{h.c}{\lambda}[/tex]

Energy of photons for ultraviolet light:

[tex]E_{uv}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{3.5\times 10^{-7}}[/tex]

[tex]E_{uv}=5.6794\times 10^{-19}\ J[/tex]

Since the energy of photons is inversely proportional to the wavelength of the light hence the photon-energy of ultraviolet light is double of the photon-energy of the red light as the wavelength of the red light is twice to that of ultraviolet light.

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

               w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

           w = 2π f = 2π / T

            2π / T = √ k / m

            k₀ / m₀ = (2π / T)²

            k₀ / m₀ = (2π / 3.0)²

            k₀ / m₀ = 4.3865

The period on the new planet is

          2π / T = √ k / m

           T = 2π √ m / k

In this case the amounts are

           m = 6 m₀

           k = 10 k₀

We replace

            T = 2π√6m₀ / 10k₀

            T = 2π √6/10 √m₀ / k₀

            T = 2π √ 0.6  √1 / 4.3865

            T = 3.23 s

Answer

IF the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would _increase_ and the wavelength of peak emission would shift toward _shorter_ wavelengths.

Explanation:

The Energy of radiation emitted by the earth varies directly with the average surface temperature of the earth and inversely with the wavelength of emissions.

E = hv/λ

That is, E = k/λ

Therefore the most peak emissions (highest energies) would have shorter wavelengths.

Complete question:

if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would __________ and the wavelength of peak emission would shift toward __________ wavelengths.

Answer:

if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would Increase and the wavelength of peak emission would shift towards Shorter wavelengths.

Explanation:

Stefan-Boltzmann law, a fundamental law of physics, explains the relationship between an object's temperature and the amount of radiation that it emits. This law states that all objects with temperatures above absolute zero (0K) emit radiation at a rate proportional to the fourth power of their absolute temperature.

Expressed mathematically as; E = σT⁴

From this formula above, temperature is directly proportional to amount of radiation emitted.

Also, Energy of emitted radiation can be related to wavelength in the expression below

E =hc/λ

Where;

E is the energy of the emitted radiation

h is Planck's constant

c is the speed of light

λ is the wavelength of the emitted radiation

From the formula above, Energy of the emitted radiation is inversely proportional to the wavelength of the emitted rays.

Answer:

a) time needed to make a round trip = DV/v2 - u2

b) time needed to make a round trip = D/root ( v2 -u2)

Explanation:

The detailed explanation is as shown in the attachment

For a round trip moving upstream and downstream, the formula for the total time needed is D / (v - u) + D / (v + u). But, if the boat moves directly across the river and back, the total time needed is D / sqrt(v² - u²). This involves the concept of relative velocity.

First, we need to understand the concept of relative velocity here. When the boat travels upstream, it moves against the current, so its net speed is (v - u). Downstream, it moves with the current at a net speed of (v + u).

A) The total time for a round trip moving upstream and downstream is the total distance divided by the speed in each direction. The time to travel a distance D upstream is D / (v - u), and the time to travel the same distance downstream is D / (v + u). So, the formula for the total time needed is D / (v - u) + D / (v + u).

B) If the boat is going directly across the river and back, it is moving perpendicular to the current. The boat makes the round trip at a constant speed of sqrt(v² - u²). Hence, the time needed for a round trip of total distance D is D / sqrt(v² - u²).

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Answer:

85 N

Explanation:

Given that crate mass = 20kg

Distance = 6m

Time = 3 seconds

Coefficient of kinetic friction = 0.3

We begin by calculating for acceleration

Which was gotten as 1.33 m/s sq

SEE THE ATTACHEMENT FOR DETAILS

To determine the tension developed in the cable, we consider forces acting on the crate, including the tension in the cable and the force of kinetic friction. We use Newton's second law of motion and equations for tension and force of kinetic friction to calculate the tension. By substituting values into the equations, we can solve for the tension.

To determine the tension developed in the cable, we need to consider the forces acting on the crate. Since the crate is moving with constant acceleration, we can use Newton's second law of motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the forces acting on the crate are the tension in the cable and the force of kinetic friction. The tension in the cable can be calculated using the equation:

Tension = mass of the crate * acceleration + force of kinetic friction

Given that the mass of the crate is 20 kg and the acceleration is the change in velocity divided by the time taken (6 m / 3 s = 2 m/s), we can calculate the tension using the equation:

Tension = 20 kg * 2 m/s^2 + force of kinetic friction

We also need to determine the force of kinetic friction. The force of kinetic friction can be calculated using the equation:

Force of kinetic friction = coefficient of kinetic friction * normal force

The normal force is equal to the weight of the crate, which can be calculated using the equation:

Normal force = mass of the crate * gravitational acceleration

Given that the coefficient of kinetic friction is 0.3 and the gravitational acceleration is 9.8 m/s^2, we can calculate the force of kinetic friction using the equation:

Force of kinetic friction = 0.3 * (20 kg * 9.8 m/s^2)

Now we can substitute this value back into the equation for tension:

Tension = 20 kg * 2 m/s^2 + (0.3 * (20 kg * 9.8 m/s^2))

Solving this equation will give us the tension developed in the cable.

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Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

(c) ΔU = 5.204nJ

Explanation:

a)

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

Answer:

a. 20652 m

b. 1239120 m

c. 0.688 m

Explanation:

First we have to find the speed of sound in air of temperature 20°C.

The speed of sound in a medium is dependent on the temperature of that medium. At any given temperature, the speed of sound is given as:

[tex]v(T) = v(T_{0}) + 0.61T[/tex]

where

v(T) = Speed of sound at any temperature T;

[tex]v(T_{0})[/tex] = Speed of sound at 0°C;

T = temperature of the medium.

The speed of sound in air of 0°C is 332 m/s, hence, in air of temperature 20°C, speed will be:

[tex]v(20) = 332 + (0.61 * 20)\\\\v(20) = 332 + 12.2\\\\v(20) = 344.2 m/s[/tex]

a. Distance traveled in 60 seconds:

distance = speed * time

distance = 344.2 * 60

distance = 20652 m

b. Distance traveled in 1 hour:

1 hour in seconds = 1 * 60 * 60 = 3600 seconds

distance = 344.2 * 3600

distance = 1239120 m

c. Distance traveled in 0.002 second:

distance = 344.2 * 0.002

distance = 0.688 m

Final answer:

The force exerted by rope R1 on block 1 is equal to the tension in that rope, and if both blocks have the same acceleration and there are no other external forces, this tension is also equal to the force R1 exerts on block 2. However, the force applied through rope R2 on block 2 may be larger because it must account for the total mass and acceleration of both blocks.

Explanation:

When two blocks are connected by a rope and a force is applied to one of the blocks (via a second rope), the tensions in the ropes are subjects of interest in Physics, specifically classical mechanics. In this scenario, the force applied to rope R2 is used to accelerate both blocks 1 and 2. The tension in rope R1, which connects blocks 1 and 2, should be analyzed using a free-body diagram for each block.

Rope R1 exerts a force on block 1, and rope R2 exerts a force on block 2. The force exerted by rope R2 must be sufficient not only to accelerate block 2 but also to account for the force needed to accelerate block 1. Therefore, the force exerted by R2 on block 2 is equal to the sum of the forces necessary to accelerate both blocks. This total force must be greater than or equal to the force that R1 exerts on block 1, as rope R1 only has to provide the force needed to accelerate block 1. However, if we assume the acceleration of the two blocks is the same and the system is not experiencing any external forces other than the applied force, the tension in rope R1 will be equal to the force R1 exerts on block 1 because the massless ropes can only transmit the force equally along their lengths.

Therefore, it can be concluded that the magnitude of the force exerted by rope R1 on block 1 is equal to the magnitude of the tension on block 2 by rope R1, but the magnitude of the applied force through R2 on block 2 could be larger as it has to account for the combined mass of block 1 and block 2.

Answer:

19 800 lbm of carbon dioxide more.

Explanation:

Taurus

Amount of gasoline used in 5 years = 650 ga/year * 5 years = 3250 ga

amount of carbon dioxide released = 19.8 lbm/ga * 3250 ga = 64 350 lbm

Explorer

Amount of gasoline used in 5 years = 850 ga/year * 5 years = 4250 ga

amount of carbon dioxide released = 19.8 lbm/ga * 4250 ga = 84 150 lbm

Extra amount of CO2 released = 84 150 lbm - 64 350 lbm = 19 800 lbm

The extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.

Taurus  

Amount of gasoline used in 5 years =

[tex]\bold {650\ gallon/year \times 5 = 3250\ gallon}[/tex]

Amount of carbon dioxide released =

[tex]\bold {19.8\ lbm/gallon \times 3250 = 64 350 lbm}[/tex]

Explorer  

Amount of gasoline used in 5 years =

[tex]\bold {850\ gallons/year \times 5 = 4250\ gallons }[/tex]

Amount of carbon dioxide released =[tex]\bold { 19.8\ lbm/gallons \times 4250 gallons = 84 150\ lbm}[/tex]

Extra amount of  [tex]\bold {CO_2}[/tex] released =

[tex]\bold {= 84 150 - 64 350 = 19 800\ lbm}[/tex]

Therefore, the extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.

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Answer:

The speed of Sam at the bottom is 7.19 m/s.

Explanation:

Given that,

Mass of Sam = 79 kg

Height = 11 m

Length = 120 m

Coefficient of kinetic friction = 0.07

Suppose, an object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, micro-kilometer , is small enough that the object will slide down the slope if given a very small push to get it started.

We need to calculate the speed at the bottom

Using conservation of energy

[tex]P.E=K.E+\text{energy lost of friction}[/tex]

[tex]mgh=\dfrac{1}{2}mv^2+\mu mg(\sqrt{L^2-h^2})[/tex]

[tex]v^2=2gh-2\mu g(\sqrt{L^2-h^2}[/tex]

[tex]v=\sqrt{2gh-2\mu g(\sqrt{L^2-h^2}}[/tex]

Where, m = mass

h = height

L= length

v = speed

g = acceleration due to gravity

Put the value into the formula

[tex]v=\sqrt{2\times9.8\times11-2\times0.07\times9.8(\sqrt{120^2-11^2})}[/tex]

[tex]v=7.19\ m/s[/tex]

Hence, The speed of Sam at the bottom is 7.19 m/s.

Velocity is the rate of change of position. Sam's velocity at the bottom of the slope with a height of 11 m is 13.8844 m/s.

Velocity is the rate of change of position of an object with respect to time.

[tex]v = \dfrac{ds}{dt}[/tex]

We know that at the topmost height, the weight of Sam will act as potential energy, while during skiing down this potential energy will be partially converted to kinetic energy and part will be converted to heat or can say will be lost due to the friction, therefore,

Potential Energy = Kinetic energy + Energy loss due to the friction,

[tex]mgh = \frac{1}{2}mv^2 + \mu gh\sqrt{L^2+H^2}\\\\mgh - \mu gh\sqrt{L^2+H^2} = \frac{1}{2}mv^2\\\\2mgh - 2\mu gh\sqrt{L^2+H^2} = mv^2\\\\2gh(m - \mu \sqrt{L^2+H^2}) = mv^2\\\\v^2 = \dfrac{2gh}{m}(m - \mu \sqrt{L^2+H^2})[/tex]

Substitute the values,

Mass, m = 79 kg

Height, H = 11 m

Length, L = 120 m

Coefficient of kinetic friction, μ = 0.07

Acceleration due to gravity, g = 9.81 m/s²

[tex]v^2 = \dfrac{2 \times 9.81 \times 11}{79}[79-0.07\sqrt{120^2+11^2}]\\\\v^2 = 215.82 - 23.0442\\\\v = 13.8844\rm\ m/s[/tex]

Hence, Sam's velocity at the bottom of the slope with a height of 11 m is 13.8844 m/s.

Learn more about Velocity:

https://brainly.com/question/862972

Answer:

C. At the instant the ball reaches its highest point.

Explanation:

When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"

Remember, F = ma = m(v/t)

Since v = 0 at maximum height

F = m(0/t)

F = 0N

This shows that the force acting on the body is zero at the maximum height.

Final answer:

Alice and Tom will both have the same speed upon reaching the water because gravity accelerates them both at the same rate for their vertical descent. Alice's initial horizontal velocity does not affect her vertical downward acceleration.

Explanation:

The correct answer is C) they will both have the same speed. This is because their vertical descent is solely affected by gravity, which accelerates all objects at the same rate (approximately 9.8 m/s2) regardless of their horizontal velocity component. Alice's initial horizontal velocity of 25 m/s contributes only to her horizontal movement and does not affect the rate at which she falls vertically due to gravity. Consequently, both Alice and Tom will have the same vertical speed when they reach the lake. When you consider both vertical and horizontal components for Alice, her overall splashdown speed will be greater than Tom's; however, the question asks about just the speed. Since only the vertical descent is mentioned, we infer that it's the vertical component being considered. Therefore, their speeds, considering just their vertical motion under gravity, are the same.

Answer: Option (b) is the correct answer.

Explanation:

Heat is defined as the energy obtained by the molecules of an object that travels from a hotter body to a colder body.

For example, a metal spoon placed in a cup full of hot tea. Then transfer of heat will take place from the tea to the spoon.

And, this transfer of heat will continue till the temperature of both the objects become equal.

Thus, we can conclude that heat is the energy that moves from a hotter object to a colder object.

Answer: The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is positive.

Explanation:

The human activity of burning fossil fuels inevitably contributes to the increase in carbon dioxide and other gases in the Earth's atmosphere. As a result of this occurrence, a natural greenhouse appears. The greenhouse effect is the process of warming the Earth's surface and the lower layers of the atmosphere, resulting from the leakage of heat radiation.

Answer:

C

Explanation:

earth science climate change unit test

Answer: D) occurs when water is evaporated.

Explanation:

A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.

A chemical change is defined as a change in which a change in chemical composition takes place. A new substance is formed in these reactions.

a).  iron rusts :the chemical reaction occurs by combination of iron with oxygen , thus a chemical change.

b) sugar is heated into caramel. the chemical reaction occurs by decomposition , thus a chemical change.

c) glucose is converted into energy within your cells: the chemical reaction occurs by decomposition , thus a chemical change.

d) when water is evaporated : Only the state changes and thus a physical change and cannot be reversed.

e) when propane is burned for heat: the chemical reaction as propane combines with oxygen to form carbon dioxide and water , thus a chemical change.

Answer:

3 units

Solution:

V=539 cubic units

Square base, with edge a=7 units

Slanted edge length: s=14 units

V=Ab h

Ab=49 square units

539 cubic units = (49 square units) h

h= 11 units

s-h=14 units-11 units

s-h=3 units

Answer:

3 units

Explanation :

Volume of an oblique rectangular prism with a square base = A×B×h

Where h = perpendicular height

From the question, Volume of an oblique rectangular prism with a square base = 539 cubic units

We were asked from the question to find how many units longer the slanted edge length of the prism, 14 is compared to its perpendicular height.

The first step is : Find the perpendicular height

Edges A = 7 units

Edges B = 7 units

Perpendicular height ?

Hence,

539 = 7 × 7 × h

539 = 49h

h = 539 ÷ 49

h = 11 units

Therefore, perpendicular height of the prism = 11 units

To find how many units longer, we would subtract the perpendicular height of the prism from the slanted edge length of the prism

= 14 units - 11 units

= 3 units .

Therefore the slanted edge length of the prism , 14, is 3 units longer compared to its perpendicular height.

Answer:

Bulbs in series will be dim

Explanation:

The voltage remain constant in a series circuit, thus bulbs in series circuit will get equally distributed voltage which is not equal to the voltage across the circuit or the voltage supplied

In case of parallel circuit, the voltage across the bulb is the same as the voltage across the circuit due to which bulbs will be brighter in case of parallel circuit