A 12 A fuse is placed in a parallel circuit that has two branches. 8 A flows in branch 1 and 6 A flows in branch 2. This fuse
1. will blow because the total current in this circuit is 14 A which is greater than 12 A.
2. will blow because a 12 A fuse only allows 6 A to flow in each branch.
3. will not blow because each branch is less than 12 A.
4. will not blow because the average current is 7 A which is less than 12 A.

Answer:

speed after one riveting operation = 230.5 rpm

mass moment of inertia = 9.90 kg

Explanation:

given data

torque energy  = 2.5 kW = 2500 W

mass = 125 kg

radius = 700 mm

energy = 1000 J

Speed of the flywheel N = 240 rpm

solution

we know here Speed of the flywheel N so here angular speed ω

ω = [tex]\frac{2\pi N}{60}[/tex]   .....................1

ω = [tex]\frac{2\pi 240}{60}[/tex]

ω = 25.13 rad/s

so here

change in energy is

ΔE = E1 - E2    ..............2

ΔE = 2500 - 1000

ΔE = 1500 J/sec

and

I = mr²    .........3

I = 125 × 0.7²

I = 61.25 kg-m²

and ΔE is express as here

ΔE = 0.5 × I × (ω² - ω1² )      ........4

put here value and we get

1500 =  0.5 × 61.25 × (25.13² - ω1² )

ω1 = 24.13 rad /s

and

reduction in speed after one riveting operation will be

N = [tex]\frac{24.13\times 60 }{2\pi }[/tex]

speed after one riveting operation = 230.5 rpm

and

for 35 rpm

ω1 = [tex]\frac{2\pi 35}{60}[/tex]

ω1 = 3.65

so ΔE  will be here

ΔE = 0.5 × mr² × (ω² - ω1² )    ....................5

put here value and we get m

1500 = 0.5 × m (0.7)² × (25.13² - 3.65² )

solve it we get

mass moment of inertia = 9.90 kg

Answer:

Explanation:

a ) angular frequency ω = [tex]\sqrt{\frac{k}{m} }[/tex]

k is spring constant and m is mass attached

ω = [tex]\sqrt{\frac{20}{1.5} }[/tex]

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

The frequency of the oscillation is 0.58 Hz and the displacement x varies as a cosine function. The maximum speed is 0.366 m/s and maximum acceleration is 0.427 m/s². The kinetic and potential energies are equal when the mass is 0.0707m away from the equilibrium point.

(a) The frequency of the oscillations is given by the formula f = 1/2π √(k/m), where k is the spring constant and m is the mass. Substituting the given values we get f = 1/2π √(20.0 N/m / 1.5 kg) = 0.58 Hz. The displacement as a function of time is given by the equation x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. Given the mass was released from rest, the phase angle is 0 and the equation becomes x(t) = (0.10 m) cos(2πft).

(b) The maximum speed is given by the formula vmax = Aω, substituting the values, vmax = (0.10 m)(2πf) = 0.366 m/s. The maximum acceleration is given by the formula amax = Aω² which is amax = (0.10 m)(2πf)² = 0.427 m/s².

(c) The kinetic energy and potential energy of the system are the same when the displacement is equal to the amplitude divided by √2: x = A/√2, that is x = 0.10m /√2 = 0.0707 m from the equilibrium point.

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Answer:

b

Explanation:

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.

Answer:

The correct answer is :

A . Giant star

B . proto star

F .main sequence star

Explanation:

hope this helps

Answer:

If the frequency is further increased (in a gradual manner), the amplitude of the oscillation of the ball would decrease.

Explanation:

As the frequency of driven force increases, the amplitude of the oscillation of the ball increases. This type of oscillation is referred to as forced oscillation.

At a particular higher frequency, the amplitude of oscillation becomes maximum. This frequency is referred to as the resonant frequency.

If the frequency is further increased (in a gradual manner), the amplitude of the oscillation of the ball would decrease.

Answer:

$2018.016

Explanation:

Applying,

P = I²R.............. Equation 1

Where P = power, drawn by the oven, I = current drawn by the oven, R = resistance of the oven.

Given: I = 10 A, R = 2002 Ω

Substitute this values into equation 1

P = 10²(2002)

P = 100(2002)

P = 200200 W.

P = 200.2 kW

If the electric oven works 3 hours daily,

Then the total time it works in a week = 3×7 = 21 hours.

E(kWh) = P(kW)×t(h)

E = 200.2×21

E = 4204.2 kWh.

If 1 kWh of energy cost $0.48,

Then, 4204.2 kWh =  $4204.2(0.48) =  $2018.016

Answer:[tex]\omega _f=1.185\ rad/s[/tex]

Explanation:

Given

mass of objects [tex]m=5\ kg[/tex]

Initially mass is at [tex]r=0.9\ m[/tex]

Initial angular speed [tex]\omega_i=0.66\ rad/s[/tex]

Moment of inertia of student and  stool is [tex]I_s=8\ kg-m^2[/tex]

Finally masses are at a distance of [tex]r_f=0.31\ m[/tex] from axis

[tex]I_i=I_p+I_m[/tex]

[tex]I_i=8+2\times 5\times (0.9)^2[/tex]

[tex]I_i=16.1\ kg-m^2[/tex]

Final moment of inertia of the system

[tex]I_f=I_s+I_m[/tex]

[tex]I_f=8+2\times 5\times (0.31)^2[/tex]

[tex]I_f=8+0.961=8.961\ kg-m^2[/tex]

As there is no external torque therefore moment of inertia is conserved

[tex]I_i\omega _i=I_f\omega _f[/tex]

[tex]\omega _f=\frac{16.1}{8.96}\times 0.66[/tex]

[tex]\omega _f=1.796\times 0.66[/tex]

[tex]\omega _f=1.185\ rad/s[/tex]

Final answer:

When the student pulls the objects closer, his angular speed increases to conserve angular momentum.

Explanation:

Angular momentum is conserved in this scenario. Initially, the student with the objects has a certain angular speed and moment of inertia. When he pulls the objects closer, his moment of inertia decreases, resulting in an increase in angular speed to conserve angular momentum.

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e           I = [tex]\frac{Q}{t}[/tex] ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

            Q = I × t

                = 1.5 × 2

               = 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

Answer:

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

Explanation:

Kinetic energy is

            K = ½ m v²

the speed of the expensive we can find it r

            v² = v₀² + 2 a x

we can find acceleration with Newton's second law

            F = m a

             a = F / m

             F= cte

substitute in the velocity equation

           v² = v₀² + 2 F/m  x

let's substitute in the kinetic energize equation

         K = ½ m (v₀² + 2 F/m  x)

         K = ½ m v₀²  + f x

we see that the kinetic energy depends on two tomines

in January in these systems the force for launching is constant, which is why decreasing the mass increases the speed of the vehicle and therefore increases the kinetic energy

As the launch speed increases the initial energy increases quadratically

we see that to increase the energy of the expensive we must increase the launch speed, since it increases quadratically

Answer:

Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy ( D )

Explanation:

when conducting plate swings through a magnetic field, energy is transferred from mechanical energy to thermal energy and then back to mechanical energy and this is due to the induction of eddy currents during the process of that the conducting plate swings through the magnetic field.

An eddy current is induced in conductor when the magnetic field in which the conductor is located varies.

Answer:

Energy is transferred from mechanical energy to thermal energy. (A)

Explanation:

As the conducting plate is swinging, we know it posseses some form of mechanical energy (either K or U). Because the plate is conducting, charge is transfered, yielding energy transfer to thermal.

Answer:

The period is  [tex]T = 8.27 \ sec[/tex]

Explanation:

From the question we are told that

   The extension of the spring is  [tex]e_1 = 33 \ m[/tex]

   The  first weight  applied is  [tex]F_1 = 19 N[/tex]

     The second weight applied is  [tex]F_2 = W[/tex]

     The second extension is [tex]e_2 = 17 \ m[/tex]

The spring constant of the spring is mathematically evaluated as

         [tex]k = \frac{F_1}{e_1 }[/tex]

substituting values

       [tex]k = \frac{19}{33 }[/tex]

       [tex]k = 0.576[/tex]

We are told that

         19 N extended the spring to 33 m    

Then W N  will extended it by  17 m

Therefore     [tex]W = \frac{19 * 17}{33}[/tex]

                    [tex]W = 9.788 \ N[/tex]

Generally the period of the oscillation is mathematically represented as

           [tex]T = 2 \pi \sqrt{\frac{M}{K} }[/tex]

where M is the mass of the W which is mathematically evaluated as

         [tex]M = \frac{9.788}{9.8}[/tex]

          [tex]M = 1.0 \ kg[/tex]

substituting values

        [tex]T = 2 \pi \sqrt{\frac{1}{0.576} }[/tex]

       [tex]T = 8.27 \ sec[/tex]

The force required to stretch a spring is directly proportional to the amount the spring is stretched. The force constant can be calculated using the equation F = kx. By substituting the given values into the equation, we can find the unknown weight W.

The force required to stretch a spring is directly proportional to the amount the spring is stretched. This can be represented by the equation F = kx, where F is the force, k is the force constant, and x is the amount the spring is stretched. In this scenario, we are given that a 19 N weight stretches the spring by 33 m at equilibrium.

To find the force constant, we can use the equation:

k = F / x = 19 N / 33 m = 0.58 N/m

Next, we need to find the unknown weight W that will stretch the spring to a new equilibrium position 17 m below the position with no weight attached. Using the force constant we calculated, we can use the equation F = kx to find the force required:

F = kx = (0.58 N/m)(17 m) = 9.86 N

Therefore, the unknown weight W that will stretch the spring to the new equilibrium position is 9.86 N.

Answer:

Explanation:

a ) Let angle of incidence and angle of refraction be i and r respectively .

submarine is 300 m away from the shore . The point where laser strikes the surface of sea is 90 m horizontally away .

Tan r = 90 / 120

= 3 / 4

.75

r = 37 degree

c ) sini / sin37 = 1.333

sini = .8

i = 53 degree

Tan 53 = 210 / h , h is height of the building .

h = 210 / tan 53

= 158 m

Answer:

99 dB

Explanation:

To find the new sound intensity level you calculate first the initial intensity by using the following formula:

[tex]\beta=10log(\frac{I}{I_o})\\\\10^{\frac{\beta}{10}}=10^{log(\frac{I}{I_o})}\\\\10^{\frac{\beta}{10}}=\frac{I}{I_o}\\\\I=I_o10^{\frac{\beta}{10}}[/tex]

where β is the sound level of 79dB and Io is the hearing threshold of 10^-12 W/m^2. By replacing you obtain:

[tex]I=(10^{-12}W/m^2)10^{\frac{79}{10}}=7.94*10^{-5}W/m^2[/tex]

The new sound intensity level is given by:

[tex]\beta'=10log(\frac{100I}{I_o})=10log(\frac{100(7.94*10^{-5}W/m^2)}{10^{-12}W/m^2})\\\\\beta'=99\ dB[/tex]

hence, the answer is 99 dB

Answer:1V

Explanation: simply put it

V=W/q

=1/1

=1V

The potential difference between these two point is 1V which can be determined from the equation of potential difference that involves work done and charge.

Potential difference is the work done per unit charge. A potential difference of 1 V means that 1 joule of work is done per coulomb of charge. Potential difference in a circuit is measured using a voltmeter which is placed in parallel with the component of interest in the circuit.

Given:

Work = 1.0 Joule

Charge = 1.0 coulomb

V=W/q

V=1/1

V=1 V

Thus, the potential difference is one Volts.

Find more information about Potential difference here:

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Answer:

Explanation:

12.0 kv primary voltage

315 kv secondary voltage ( converted voltage ) V1 or Vo

v2 (Vn)= 730 kv new secondary voltage

a) Ratio of turns in 730 kv to turns in 315 kv

[tex]\frac{Vn}{Vo} = \frac{Nn}{No}[/tex] = [tex]\frac{730}{315}[/tex]  therefore the ratio of turns = 2.317 ≈ 2.32

B) ratio of the new current output to the old current output for the same power input to the transformer

since the power input is the same

[tex]\frac{In}{Io} = \frac{\frac{Vp}{Vn} }{\frac{Vp}{Vo} }[/tex]     equation 1

Vp = primary voltage, Vo = old secondary voltage, Vn = new secondary voltage, In = new secondary current, Io = old secondary current

therefore equation 1 becomes

[tex]\frac{In}{Io} = \frac{Vo}{Vn}[/tex] =  315 / 730 = 0.43

Final answer:

The ratio of turns in the new secondary compared to the old secondary is 2.32, and the ratio of the new current output to the old current output for the same power input to the transformer is approximately 0.432.

Explanation:

Transformer Coil Turn Ratio and Current Output Ratio

In a transformer, the ratio of the voltage across the secondary coil to the voltage across the primary coil is equal to the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. For the same power input to a transformer, the current output will be inversely proportional to the voltage output. Thus, when the voltage output is increased, the current output decreases proportionally.

(a) Ratio of Turns in the New Secondary to the Old Secondary

To find the ratio of turns in the new secondary compared to the old secondary, we use the ratio of the new voltage to the old voltage:

New ratio of turns (N2) / Old ratio of turns (N1) = V2 / V1 = 730 kV / 315 kV = 2.32.

(b) Ratio of the New Current Output to the Old Current Output

For the same power input, P = VI, where P is the power, V is the voltage and I is the current. Keeping the power constant and increasing the voltage output results in a decrease in current. Therefore, the ratio of the new current to the old current is inversely proportional to the ratio of the new voltage to the old voltage:

New current (I2) / Old current (I1) = V1 / V2 = 315 kV / 730 kV = 0.432.

Answer:

The initially known quantity, together with the wavelength, that is sufficient to calculate the stopping potential for electrons from the surface of a metal is called the WORK FUNCTION.

Explanation:

The stopping potential is defined as the potential that is required to stop electrons from being ejected from the surface of a metal when light with energy greater than the metal's work function/work potential is incident on the metal.

Given that light is known to be made up of photons, which carry energy in packets according to the frequencies of the light.

The photoelectric phenomenon explains that when light of a certain frequency that corresponds to an energy level that is higher than a metal's work function is incident on a metal, it will lead to electrons being ejected from the surface of the metal. The energy of the ejected electrons is then proportional to the difference between the energy level of the photons and the metal's work function.

Basically, it is the excess energy after overcoming the work function that rejects the electrons.

So, to prevent this excess energy from ejecting electrons from a metal's surface, an energy thay matches this excess must be in place to stop electrons from coming out. This energy/potential required to stop the ejection of electrons, is called the stopping potential.

The stopping potential is given as

eV₀ = hf - ϕ

The stopping potential (eV₀) them depends on the hf and the ϕ.

hf is the energy of the photons, where h is Planck's constant and f is the photons' frequency which is further given as

f = (c/λ)

c = speed of light (speed of the photons)

λ = wavelength of the photons.

The other quantity, ϕ, is the metal's work function; the amount of energy needed to be overcome by the photons before ejection of electrons is possible. It is the minimum energy that the light photoms must possess to even stand a chance of being able to eject electrons from a metal's surface.

So, the stopping potential is the difference between the energy of the photons (obtained using the photons' frequency, wavelength and/or speed) and the metal's work function.

Hope this Helps!!!!

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

[tex]x=x_{0}+V t \\ x=\frac{1}{2}+V t[/tex]

The distance between the police car and the intersection is,

[tex]y=y_{0}+V t[/tex]

[tex]y=\frac{1}{2}-40 t[/tex]

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

[tex]z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })[/tex]

[tex]z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)[/tex]

The rate of change is,

[tex]2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)[/tex]

[tex]2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots[/tex]

Now finding [tex]z[/tex] when [tex]t=0,[/tex] from (1) we have

[tex]z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}[/tex]

[tex]z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071[/tex]

The officer's radar gun indicates 25 mph pointed at the other car then, [tex]\frac{d z}{d t}=25[/tex] when [tex]t=0,[/tex] from

From (2) we get

[tex]2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)[/tex]

[tex]2(0.7071)(25)=V+2 V^{2}(0)-40[/tex]

[tex]35.36=V-40[/tex]

[tex]V=35.36+40=75.36[/tex]

Hence the speed of the car is [tex]75.36 mph[/tex]

To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.

To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.

To calculate the speed of the other car, we can use the Pythagorean theorem. The police car is moving in a straight line, so the relative velocity is the hypotenuse of a right triangle. The northward velocity of the police car is one side of the triangle, and the eastward velocity of the other car is the other side. Using the equation a² + b² = c², where a is the northward velocity, b is the eastward velocity, and c is the relative velocity, we can solve for b. Rearranging the equation, we get b = √(c² - a²). Plugging in the known values, we have b = √(25² - 40²) = √(625 - 1600) = √(-975)

The square root of a negative number is not a real number, so it is not possible to determine the speed of the other car using only the given information.

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Answer:

I would say A since it's important part of the instructions.

Explanation:

Answer:

A

Explanation:

Answer:

U = 5.37*10^33 J

Explanation:

The gravitational potential energy between two bodies is given by:

[tex]U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}[/tex]

G: Cavendish's constant = 6.67*10^-11 m^3/kg.s

For three bodies the total gravitational potential energy is:

[tex]U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}][/tex]

BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:

[tex]U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J[/tex]

hence, the total gravitational energy is 5.37*10^33 J

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

[tex]m\lambda=asin\theta[/tex]

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

[tex]y=\frac{m\lambda D}{a}[/tex]

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

[tex]w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm[/tex]

b) the width of first maximum is y2-y1:

[tex]w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm[/tex]

c) and for the second maximum:

[tex]w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm[/tex]

Answer: A monochromatic light source is one that consists of many colors

Explanation:

Monochromatic comes from "mono - one, chromatic - color", so a  monochromatic wave has only one wavelength, this means that it has only one color (while it may have different shades or tones). So the correct option is the second one "A monochromatic light source is one that consists of many colors "

The other 3 statements are true.

The incorrect statement is that 'A monochromatic light source is one that consists of many colors', because a monochromatic light source indeed emits light of a single wavelength or color. The other claims regarding phase shift on reflection, and the location of highest intensity in single and double slit interference patterns are accurate.

The statement that is NOT true among the given options is: 'A monochromatic light source is one that consists of many colors'. In fact, a monochromatic light source emits light of a single wavelength or color.

Further addressing the related concepts: When a light wave encounters a boundary where it must reflect off a surface with a larger index of refraction, the phase of the wave indeed flips by pi. This is an important concept in understanding wave reflection and refraction.

Regarding the intensity of the central maximum in both single and double slit interference patterns, it is accurate to say that the intensity of the center maximum is the most significant. This consequence is a direct outcome of the interference pattern produced within these experiments. In the central maximum, most wave paths align constructively, resulting in the highest intensity.

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Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

[tex]H_{nom}=2hp=1491.4W[/tex]

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

[tex]w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m[/tex]

[tex]V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s[/tex]

[tex]F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N[/tex]

[tex](F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N[/tex]

[tex]T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm[/tex]

[tex]F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N[/tex]

[tex]F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N[/tex]

b)

[tex]H_a=1491*1.25=1863.75W[/tex]

[tex]n_f_s=\frac{H_a}{H_{nom}K_S }=1[/tex]

dip = [tex]\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm[/tex]

A) The values of  Fc, Fi, ( F1)a  and F2

B ) The values of  Ha, nfs  and belt length

Given data :

Angular velocity ratio of large pulley = 0.5

width of polyamide ( b ) = 6 inches = 0.1524 m

pulley width ( d ) = 2 inches = 0.0508 m

center to center distance ( L ) = 9 ft =  2.7432 m

angular speed of small pulley ( n ) = 1750 rev/min

power of small pulley ( Hnom ) = 2 hp = 1491.4 Watts

Ks = 1.25

γ = 9500 N/m³

g = 9.81 m/s²

t = 0.0013 mm

A) Determine the values of Fc, Fi, F1a and F2

w = γ * b * t

   = 9500 * 0.1524 * 0.0013 = 1.88 N/m

V = [tex]\frac{\pi dn}{60}[/tex] = ( π * 0.0508 * 1750 ) / 60 = 4.65 m/s  

T = 10.17 Nm ( calculated value )

i) Fc = [tex]\frac{wV^2}{g}[/tex]  ---- ( 1 )

Insert values into equation ( 1 )

Fc = ( 1.88 * 4.65² ) / 9.81

    = 4.15 N

ii) Fi = [tex]\frac{(F1)a+F2 }{2}[/tex]  - Fc ----- ( 2 )

where ( F1 )a = 0.1524 * 6000 * 0.7 * 1 = 640 N

             F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex] =  640 - [tex]\frac{2 *10.17}{0.0508}[/tex]  = 239.6 N

back to equation ( 2 )

Fi = [ ( 640 + 239.6 ) / 2 ] - Fc

   =  435.65 N

iii) ( F1 )a = b * Fa * Cp * Cv

              = 0.1524 * 6000 * 0.7 * 1  = 640 N

iv) F2 = ( F1 )a - [tex]\frac{2T}{D}[/tex]

         = 640 - [tex]\frac{2 *10.17}{0.0508}[/tex]  = 239.6 N

B) Find the values of Ha, nfs and belt length

i) Ha = 1491 * 1.25

        = 1863.75 W

ii) nfs = [tex]\frac{Ha}{Hnom*Ks } = 1[/tex]

iii) belt length ( dip ) = [tex]\frac{L^2w}{8fi}[/tex]  

                                 = ( 2.7432 * 1.88 ) / 435.65  = 11.8 mm

Hence we can conclude that the values of the variables in the question is as listed above

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Answer:

at point F

Explanation:

To know the point in which the pendulum has the greatest potential energy you can assume that the zero reference of the gravitational energy (it is mandatory to define it) is at the bottom of the pendulum.

Then, when the pendulum reaches it maximum height in its motion the gravitational potential energy is

U = mgh

m: mass of the pendulum

g: gravitational constant

The greatest value is obtained when the pendulum reaches y=h

Furthermore, at this point the pendulum stops to come back in ts motion and then the speed is zero, and so, the kinetic energy (K=1/mv^2=0).

A) answer, at point F

Answer:

at point F

Explanation:

Answer:

a) An equation for the x-component of the electric field.

Eₓ = (-15xy³ + 5.32xy⁴z²) N/C

b) An equation for the y-component of the electric field.

Eᵧ = (-22.5x²y² + 10.64x²y³z²) N/C

c) An equation for the z-component of the electric field.

Ez = (5.32x²y⁴z) N/C

d) At (-5.0, 2.0, 1.5) m, the electric field is given as

E = (-357.6î + 2,538ĵ + 3,192ķ) N/C

Magnitude of the electric field = 4,093.7 N/C

Explanation:

The electric field is given by the negative of the gradient of the electric potential,

E = −grad V

E = - ∇V

The electric potential is given as

V(x,y,z) = 3αx²y³ - 2γx²y⁴z²

α = 2.5 V/m⁵ and γ = 1.33 V/m⁸

V(x,y,z) = 7.5x²y³ - 2.66x²y⁴z²

grad = ∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)ķ

E = -grad V = -∇V

= -[(∂V/∂x)î + (∂V/∂y)ĵ + (∂V/∂z)ķ

E = -(∂V/∂x)î - (∂V/∂y)ĵ - (∂V/∂z)ķ

E = Eₓî + Eᵧĵ + Ez ķ

a) An equation for the x-component of the electric field.

Eₓ = -(∂V/∂x) = -(∂/∂x)(V)

= -(∂/∂x)(7.5x²y³ - 2.66x²y⁴z²)

= -(15xy³ - 5.32xy⁴z²)

= (-15xy³ + 5.32xy⁴z²)

b) An equation for the y-component of the electric field.

Eᵧ = -(∂V/∂y) = -(∂/∂x)(V)

= -(∂/∂y)(7.5x²y³ - 2.66x²y⁴z²)

= -(22.5x²y² - 10.64x²y³z²)

= (-22.5x²y² + 10.64x²y³z²)

c) An equation for the z-component of the electric field.

Ez = -(∂V/∂z) = -(∂/∂x)(V)

= -(∂/∂z)(7.5x²y³ - 2.66x²y⁴z²)

= -(0 - 5.32x²y⁴z)

= (5.32x²y⁴z)

d) E = Eₓî + Eᵧĵ + Ez ķ

E = (-15xy³ + 5.32xy⁴z²)î + (-22.5x²y² + 10.64x²y³z²)ĵ + (5.32x²y⁴z) ķ

At (-5.0, 2.0, 1.5) m

x = -5 m

y = 2 m

z = 1.5 m

Eₓ = (-15xy³ + 5.32xy⁴z²)

= (-15×-5×2³) + (5.32×-5×2⁴×1.5²)

= 600 - 957.6 = -357.6

Eᵧ = (-22.5x²y² + 10.64x²y³z²)

= (-22.5×(-5)²×2²) + (10.64×(-5)²×2³×1.5²)

= -2250 + 4788 = 2538

Ez = (5.32x²y⁴z) = (5.32×(-5)²×2⁴×1.5)

= 3192

E = -357.6î + 2,538ĵ + 3,192ķ

Magnitude = /E/ = √[(-357.6)² + 2538² + 3192²]

= 4,093.6763135353 = 4,093.7 N/C

Hope this Helps!!!!

The correct answers are as follows:

(a) The equation for the x-component of the electric field is[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The equation for the y-component of the electric field is [tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The equation for the z-component of the electric field is [tex]\[ E_z = 5.32x^2y^4z \][/tex]

(d) the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb is 14907.5 N/C.

To find the components of the electric field, we need to take the negative gradient of the electric potential function V(x,y,z). The gradient of a function is a vector field whose components are the partial derivatives of the function with respect to each variable. The electric field [tex]\( \vec{E} \)[/tex] is related to the electric potential [tex]\( V \)[/tex] by the equation:

[tex]\[ \vec{E} = -\nabla V \][/tex]

where [tex]\( \nabla \)[/tex] is the gradient operator.

(a) The x-component of the electric field [tex]\( E_x \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( x \)[/tex] :

[tex]\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_x = -3α(2xy^3) + 2γ(2xy^4z^2) \][/tex]

[tex]\[ E_x = -6αxy^3 + 4γxy^4z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_x = -6(2.5)xy^3 + 4(1.33)xy^4z^2 \][/tex]

[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The y-component of the electric field [tex]\( E_y \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( y \)[/tex] :

[tex]\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex] [tex]\[ E_y = -3αx^2(3y^2) + 2γx^2(4y^3z^2) \][/tex]

[tex]\[ E_y = -9αx^2y^2 + 8γx^2y^3z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_y = -9(2.5)x^2y^2 + 8(1.33)x^2y^3z^2 \][/tex]

[tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The z-component of the electric field

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( z \)[/tex] :

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -2γx^2y^4(-2z) \][/tex]

[tex]\[ E_z = 4γx^2y^4z \][/tex]

Substituting the given value of[tex]\( γ \)[/tex] :

[tex]\[ E_z = 4(1.33)x^2y^4z \]\\ E_z = 5.32x^2y^4z \][/tex]

(d) To calculate the magnitude of the electric field at the point[tex]\( P(-5.0, 2.0, 1.5) \)[/tex] m, we first substitute [tex]\( x = -5.0 \)[/tex] m, [tex]\( y = 2.0 \)[/tex] m, and [tex]\( z = 1.5 \)[/tex] m into the equations for[tex]\( E_x \)[/tex], [tex]\( E_y \)[/tex] , and [tex]\( E_z \)[/tex] :

[tex]E_x = -15(-5.0)(2.0)^3 + 5.32(-5.0)(2.0)^4(1.5)^2 \]\\ E_x = 1500 - 5.32(-5.0)(16)(2.25) \]\\ E_x = 1500 - (-851.88) \]\\ E_x = 1500 + 851.88 \]\\ E_x = 2351.88 \text{ N/C} \][/tex]

[tex]\[ E_y = -22.5(-5.0)^2(2.0)^2 + 10.64(-5.0)^2(2.0)^3(1.5)^2 \]\\ E_y = -22.5(25)(4) + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 4752 \]\\ E_y = 2552 \text{ N/C} \][/tex]

[tex]\[ E_z = 5.32(-5.0)^2(2.0)^4(1.5) \]\\ E_z = 5.32(25)(16)(1.5) \]\\ E_z = 5.32(600) \]\\ E_z = 3192 \text{ N/C} \][/tex]

Now, the magnitude of the electric field \( \vec{E} \) is given by:

[tex]\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{(2351.88)^2 + (2552)^2 + (3192)^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{5527641.64 + 6512644 + 10180416} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{22220502.64} \][/tex]

[tex]\[ |\vec{E}| \approx 14907.5 \text{ N/C} \][/tex]

Therefore, the magnitude of the electric field at the point [tex]\( P(-5.0, 2.0, 1.5) \)[/tex]m is approximately [tex]\( 14907.5 \)[/tex] N/C.

Answer:

Negatively charged or positively charged atom is generally termed as ANION/CATION. Short Explanation: If an atom loses electrons or gains protons, it will have a net positive charge and is called a Cation. If an atom gains electrons or loses protons, it will have a net negative charge and is called an Anion.

Explanation:

Answer:

Explanation:

Given

mass of rod is M

Distance between rails is L

If the direction of magnetic field is directed upwards

If rod is moving right then,

Area enclosed is increasing therefore a Clockwise current will induce

and if rod is moving left then enclosed area is decreasing therefore an anticlockwise current will setup in the circuit

(b)Induced EMF will be

[tex]E=BvL[/tex]

where v=velocity of rod

and current is [tex]I=\frac{V}R}[/tex]

[tex]I=\frac{BvL}{R}[/tex]

(c)Force necessary to move rod with constant velocity v is

[tex]F_{net}=F-BIL=0[/tex]

[tex]F=BIL[/tex]

[tex]F=B\times \frac{BvL}{R}\times L[/tex]

[tex]F=\frac{B^2L^2v}{R}[/tex]

Answer:

Explanation:

Total time of transfer oil fuel = 2.45 x 9 = 22.05 minutes

= 22.05 x 60

= 1323 s

Total volume of ful transferred = 687 gallon

= .0037854 x 687

= 2.6 m³

radius of pipe = .5 x 1.45 inch

= .5 x 1.45 x 2.54 x 10⁻² m

r = .018415 m

cross sectional area

= π r²

a = 3.14 x .018415²

= 10.648 x 10⁻⁴ m²

If v be the velocity

volume of fuel coming out  a x v x t  , a is cross sectional area , v is velocity and t is time .

10.648 x 10⁻⁴ x v x 1323  = 2.6

v = 1.845 m / s

The velocity of the fuel through the refueling nozzle was calculated to be approximately 1.845 meters per second, using the area of the nozzle and the volume of fuel transferred over the total duration of the refuelings.

To calculate the velocity of the fuel through the nozzle during the in-flight refueling mentioned, we will need to use the formula for the volume flow rate, which is Q = A × v, where Q is the volume flow rate, A is the cross-sectional area of the nozzle, and v is the velocity of the fuel.

The cross-sectional area A of the nozzle can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the nozzle. Given the diameter of the nozzle is 1.45 inches, the radius r is half of that, which is 0.725 inches. However, we should convert this to meters to use SI units: 0.725 inches × 0.0254 meters/inch = 0.018415 meters. So, A = π × (0.018415 meters)^2 ≈ 1.066 × 10^{-3} m^2.

To find the total volume of fuel transferred, we need to convert 687 gallons to cubic meters. There are 3.78541 liters in a gallon and 1,000 liters in a cubic meter, so the volume V in cubic meters is 687 gallons × 3.78541 liters/gallon × 1 m^3/1000 liters = 2.600 cubic meters.

Given that each refueling took 2.45 minutes and there were 9 refuelings, the total time t taken for the fuel transfer in seconds is 9 transfers × 2.45 minutes/transfer × 60 seconds/minute = 1321.5 seconds.

Now we can calculate the volume flow rate Q = V / t, which equals 2.600 m^3 / 1321.5 s = 1.967 × 10^{-3} m^3/s.

Finally, we can solve for the velocity v using Q = A × v, so v = Q / A which is (1.967 × 10^{-3} m^3/s) / (1.066 × 10^{-3} m^2) = 1.845 m/s.

Thus, the velocity of the fuel through the nozzle was approximately 1.845 meters per second.

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Answer:

9.2 A

Explanation:

V = IR

115 V = I (12.5 Ω)

I = 9.2 A