A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100 rad/s2. After making 2844 revolutions, its angular speed is 140 rad/s.

Answer:

x = 54.3m (on the +ve x axis)

Explanation:

This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.

Given that acceleration, a, is:

ax(t) = - 0.032(15.0 - t)

And the initial values are:

x(t = 0) = - 14.0m

v(t = 0) = 8.7m/s

Hence,

a = - 0.032(15 - t)

a = - 0.48 + 0.032t

a = dv/dt = -0.48 + 0.032t

To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:

v = ∫dv/dt  = ∫(-0.48 + 0.032t)

∫dv = ∫(-0.48 + 0.032t)dt

(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)

Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:

=> v - 8.7 = -0.48t + 0.032t²/2

v = 8.7 - 0.48t + 0.016t²

To obtain distance, x, integrate the velocity and apply the initial values:

v = dx/dt = 8.7 - 0.48t + 0.016t²

=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)

∫dx= ∫(8.7 - 0.48t + 0.016t²)dt

(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)

Inputting the initial values t₀ = 0s, x₀ = - 14.0m:

(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3

x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0

Now that the distance, x, has been obtained, when t = 10s:

x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0

x = 87 - 24 + 5.3 - 14.0

x = 54.3m

Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).

Acceleration of a body is the change in velocity with respect to time.

The x-coordinate of the object when value of time is 10 second is 54.3 m.

Acceleration of a body is the change in velocity with respect to time.

Given information-

A small object moves along the x-axis with acceleration,

[tex]ax(t)=-0.0320(15-t)[/tex]

Initial position and initial velocity of the object is,

[tex]x_0=-14\rm m\\v_0=8.7\rm m/s[/tex]

For the velocity integrate the given equation as,

[tex]ax(t)=-0.0320(15-t)\\\dfrac{dv}{dt} =\int ({-0.48+0.032t} )\, dt\\v-v_0=-0.48t+0.032\times\dfrac{t^2}{2} \\v-8.7=-0.48t+0.016t^2\\v=0.016t^2-0.48t+8.7[/tex]

Integrate it again to find the distance of the object.

[tex]v=0.016t^2-0.48t+8.7\\\int\dfrac{dx}{dt}=\int(0.016t^2-0.48t+8.7)dt\\x-x_0=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\x-14=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\\\x=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t-14\\[/tex]

Put the value of [tex]t[/tex] as 10 seconds as,

[tex]x=0.016\times\dfrac{10^3}{3}-0.48\times\dfrac{10^2}{2}+8.7\times10-14\\[/tex]

[tex]x=54.3\rm m[/tex]

Hence the x-coordinate of the object when value of time is 10 second is 54.3 m.

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Answer:

1) 2.44 joules

2) 7.73 m/s

3) 7.6 joules

Explanation:

Kinetic energy (K) of a particle is:

[tex] K=\frac{mv^{2}}{2} [/tex] (1)

with m the mass, and v the velocity

1) Because we already now velocity on A (va) and the mass of the object we can calculate its kinetic energy:

[tex]K_{a}=\frac{mv_{a}^{2}}{2}=\frac{(0.338kg)(3.8\frac{m}{s})^{2}}{2}=2.44J [/tex]

2) Because on B we know mass and kinetic energy we should solve (1) for v and use our values to find the velocity on B:

[tex]v_{b}=\sqrt{\frac{2K_{b}}{m}}=\sqrt{\frac{2(10.1J)}{(0.338kg)}}=7.73\frac{m}{s} [/tex]

3) Work-energy theorem states that the change of kinetic energy of an object is equal to the total work done on it, so:

[tex]W=K_b-K_a=10.1J-2.44J= 7.6J [/tex]

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.

Answer:

2670.90667586 J

108.573324964 dB

Explanation:

r = Distance = 120 m

A = Area = [tex]4\pi r^2[/tex]

I = Intensity of sound = [tex]7.2\times 10^{-2}\ W/m^2[/tex]

t = Time taken = 0.205 s

[tex]I_0[/tex] = Threshold intensity = [tex]10^{-12}\ W/m^2[/tex]

Power is given by

[tex]P=IA\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2[/tex]

Energy is given by

[tex]E=Pt\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2\times 0.205\\\Rightarrow E=2670.90667586\ J[/tex]

The total amount of energy is 2670.90667586 J

Sound intensity level is given by

[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow \beta=10log\dfrac{7.2\times 10^{-2}}{10^{-12}}\\\Rightarrow \beta=108.573324964\ dB[/tex]

The sound level is 108.573324964 dB

The total amount of energy transferred away from the explosion by sound is 1.476 x 10^-2 Joules. The sound level in decibels heard by the observer is 97.50 dB.

(a) To find the total amount of energy transferred away from the explosion by sound, we can use the formula:

Energy = Intensity x Time

Substituting the given values:

Energy = (7.20 x 10-2 W/m2) x (0.205 s) = 1.476 x 10-2 Joules

(b) The sound level in decibels heard by the observer can be calculated using the formula:

Sound Level (dB) = 10 x log10(Intensity / Threshold Intensity)

Substituting the given values:

Sound Level (dB) = 10 x log10((7.20 x 10-2 W/m2) / (10-12 W/m2)) = 97.50 dB

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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: False

Explanation:

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

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:

[tex] m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr[/tex]

So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C

Explanation:

For this case we need to use the fact that the sum for all the heats involved in the system are 0, since we assume an equilibrium state.

Data given

[tex]m_p = 303 gr[/tex] mass of the porcelain cup

[tex] cp_p = 0.260 cal/g C[/tex] the specific heat for the porcelain cup

[tex] T_{ip} = T_{ic}= 71 C[/tex] initial temperature for the coffee and the porcelain cup.

[tex] V_{c}= 161 cm^3[/tex] Volume of the coffee.

We can convert this to m^3 and we got 0.000161m^3 and assuming the density fot the coffee equal to the water 1 Kg/m^3 the mass would be:

[tex] m_c = 1 kg/m^3 *0.000161 m^3 = 0.000161 kg=0.161 Kg[/tex]

[tex] Cp_{c} = 1 cal/g C[/tex] Specific heat for the coffee

[tex] m_i =?[/tex] mass of ice required

[tex] T_e= 49C[/tex] equilibrium temperature

[tex]L_f = 80 cal/g [/tex] represent the latent heat of fusin since the ice change the state to liquid.

Solution to the problem

Using this formula:

[tex] \sum_{i=1}^n Q_i = 0[/tex]

We have this:

[tex] m_p cp_p (T_e -T_{ip}) + m_{c} cp_c (T_e -T_{ic}) +m_i L_f + m_i cp_w (T_e -0) =0[/tex]

Now we can replace and we have this:

[tex] 303 gr *(0.260 cal/g C) (49-71)C + 0.161 gr*(1 cal/g C)(49-71)C +m_i [80 cal/gr+(1cal/g C)(49-0)C]=0[/tex]

And now we can solve for [tex] m_i[/tex] and we have:

[tex]-1733.16cal -3.542cal +m_i [129 cal/g]=0[/tex]

[tex]m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr[/tex]

So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C

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:

[tex]F'=2F[/tex]

Explanation:

According to Coulomb's law and assuming the objects as point charges, the magnitude of the electric force that each object exerts on the other is defined as:

[tex]F=\frac{kq_Aq_B}{d^2}[/tex]

Here k is thee Coulomb constant, [tex]q_A[/tex] and [tex]q_B[/tex] are the charges of the objects and d is the distance of separation between them. We have [tex]q'_A=2q_A[/tex]:

[tex]F'=\frac{kq'_Aq_B}{d^2}\\F'=\frac{k2q_Aq_B}{d^2}\\F'=2\frac{kq_Aq_B}{d^2}\\F'=2F[/tex]

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

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 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 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: 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.

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:

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

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. 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

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.

Answer:

Technician B is correct.

Explanation:

The given value of resistor having 3% tolerance means that the given quantity of the physical parameter can vary by 3% of what is specified. This variation can be either more or less than the quantified value.

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.

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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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:

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:

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.

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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

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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: 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

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.

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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