A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits at point B and attaches to the ship at points C and D. The two rope segments BC and BD angle away from the center of the ship at angles of ϕ = 25.0 ∘ and θ = 23.0 ∘, respectively. The tugboat pulls with a force of 3700 lb . What are the tensions TBC and TBD in the rope segments BC and BD?

Answer:

P₃₅₀ = 0.28 watt

Explanation:

First we find the resistance of the wire at 20°C:

R₀ = ρL/A

where,

ρ = resistivity = 1 x 10⁻⁶ Ωm

L = Length of wire = 100 cm = 1 m

A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²

Therefore,

R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)

R₀ = 1.27 Ω

Now, from Ohm's Law:

V = I₀R₀

where,

V = Potential Difference = ?

I₀ = Current Passing at 20°C = 0.5 A

Therefore,

V = (0.5 A)(1.27 Ω)

V = 0.64 volts

Now, we need to find the resistance at 350°C:

R₃₅₀ = R₀(1 + αΔT)

where,

R₃₅₀ = Resistance at 350°C = ?

α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹

ΔT = Difference in Temperature = 350°C - 20°C = 330°C

Therefore,

R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]

R₃₅₀ = 1.44 Ω

Now, for power at 350°C:

P₃₅₀ = VI₃₅₀

where,

P₃₅₀ = Power dissipation at 350°C = ?

V = constant potential difference = 0.64 volts

I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)

Therefore,

P₃₅₀ = V²/R₃₅₉

P₃₅₀ = (0.64 volts)²/(1.44 Ω)

P₃₅₀ = 0.28 watt

Answer:

The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.

Explanation:

Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.

Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.

When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.

Answer:

The value  of  the function is  [tex]q__{t }} = 1.878 *10^{35}[/tex]

Explanation:

From the question we are told that

     The temperature is  [tex]T = 1000 \ K[/tex]

      The volume  is  [tex]V = 1 m^3[/tex]

Generally the  transnational partition function is mathematically represented as

        [tex]q__{t }} = [\frac{2 * \pi * m * k * T }{ N_a * h} ]^{\frac{3}{2} } * V[/tex]

Where  m is the molar mass of oxygen with a constant value of  [tex]m = 32 *10^{-3} \ kg/mol[/tex]

 k  is the Boltzmann constant with a value  of  [tex]k = 1.38 *10^{-23 } \ J/K[/tex]

[tex]N_a[/tex] is the Avogadro Number with a constant value of  [tex]N_a = 6.022 *10^{23} \ atoms[/tex]

h is  the Planck's  constant  with value   [tex]h = 6.626 *10^{-34 } \ J\cdot s[/tex]

Substituting values

       [tex]q__{t }} = [\frac{2 * 3.142 * 32*10^{-3} * 1.38 *10^{-23} * 1000 }{ 6.022 *10^{23} * [6.626 *10^{-34}] ^2 }]^{\frac{3}{2} } * 1[/tex]

       [tex]q__{t }} = 1.878 *10^{35}[/tex]

Answer:

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]

The value of the spring constant is 187,500N/m

Answer:

see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Explanation:

This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5

Let's analyze the light beam path emitted by the diver.

* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º

* also the wavelength of light in a material medium changes

      λ_n =  λ / n

where  λ_n is the wavelength in the material and  λ the wavelength in the vacuum air and n the refractive index.

If we include these aspects, the constructive interference equation is

       2t = (m + ½)  λ_n

       2nt = (m + ½)  λ

let's apply this equation to our case

            λ = 2nt / (m + ½)

The incidence of replacement of the oil with respect to water is

        n = n_oil / n_water = 1.5 / 1.33

        n = 1,128

let's calculate

        λ = 2 1,128 t / (m + ½)

        λ = 2,256 t / (m + ½)

In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation

          t = 0.001 mm = 1 10⁻⁶ m

the formula remains

        λ = 2,256 10⁻⁶ / (m + ½)

Let's find what values ​​of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m

     m + ½ = 2,256 10⁻⁶ / λ

     m = 2,256 10⁻⁶ / λ - ½

light purple lan = 400 10⁻⁹m

     m = 2,256 10-6 / 400 10⁻⁹ - ½

     m = 5.64 - 0.5

     m = 5.14

     m = 5

red light  λ = 700 10⁻⁹m

      m = 2,256 1-6 / 700 10⁻⁹ - ½

      m = 3.22 - 0.5

      m = 2.72

      m = 3

we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference

Answer:

Higher than the coefficient of kinetic friction.

Explanation:

Hope it helps u..   :)

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, [tex]V_T = 60 \ m/s[/tex]

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

[tex]F_D = kV_T^2[/tex]

Where;

k is a constant

[tex]k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}[/tex]

When the new drag force is half of the original drag force;

[tex]F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s[/tex]

Therefore, the speed of the resistive force is 42.426 m/s

At terminal speed, the speed of the resistive force will be:

"42.426 m/s".

According to the question,

Skydriver's mass, m = 75 kg

Terminal velocity, [tex]V_T[/tex] = 60 m/s

Constant = k

We know the relation,

→ [tex]F_D[/tex] = k[tex]V_T^2[/tex]

here, k = [tex]\frac{F_D_1}{V_T_1^2} = \frac{F_D_2}{V_T_2^2}[/tex]

Now,

  [tex]F_D_2[/tex] = [tex]\frac{F_D_1}{2}[/tex]

  [tex]\frac{F_D_1}{V_T_1^2}= \frac{F_D_2}{V_T_2^2}[/tex]

   [tex]\frac{1}{V_T_1^2} = \frac{1}{2V_T_2^2}[/tex]

By applying cross-multiplication,

  [tex]V_T_2^2 = \sqrt{\frac{V_T_1^2}{2} }[/tex]

By substituting the above values,

  [tex]V_T_2[/tex] = 0.7071 ([tex]V_T_1[/tex])

        = 0.7071 × 60

        = 42.426 m/s

Thus the above response is correct.

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Answer:

 6.758V

Explanation:

The computation of the final speed of the defender/ball carrier mass is shown below:-

Data provided in the question

Measurement of ball = 75 kg

Right m/s = 6.5 m/s

Defender = 80 kg

Carrying running 7.0 m/s

Based on the above information, the final speed of the defender or ball carrier mass is

As we know that

Conservation of momentum is

= Ball in Kg × Right m/s + Defender in Kg × running m/s

= 75 × 6.5 + 80 × 7

= (75 + 80)V

Therefore, the Final speed = 6.758V

Answer:

[tex]F=1.8\times 10^{-3}\ N[/tex]

Explanation:

We have,

Length of wires is 11 m

Separation between wires is 0.033 m

Current in both the wires is 5.2 A

It is required to find the magnitude of force between two wires. The force between wires is given by :

[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]

So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]

Answer:

The tube should be held vertically, perpendicular to the ground.

Explanation:

As the power lines  of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its  potential.

And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.

Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = [tex]I[/tex]

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = [tex]5I[/tex]  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

therefore,

for the first case, the K.E. is given as

[tex]KE = \frac{1}{2}Iw^{2}[/tex]

and for the second case, the K.E. is given as

[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]

[tex]KE = \frac{1}{10}Iw^{2}[/tex]

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Now, let's consider moment of inertia =  I  and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia =   (five times larger)

angular speed = ω/5  (five times smaller)

The kinetic energy of a spinning body is given as:

[tex]K.E.=\frac{1}{2} I. w^2[/tex]

On substituting the values, we will get:

[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

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Answer:

By a factor of 9/4

Explanation:

Applying Coulomb's law,

F = kqq'/r²................... Equation 1

Assuming q and q' are the two point charges respectively.

Where k = coulomb's constant, r = distance between the charges.

When the point charges are 3.0 m apart,

F = kqq'/3²

F = kqq'/9.................... Equation 1

When they are moved to a new distance, 2.00 m

F' = kqq'/2²

F' = kqq'/4................. Equation 2.

Comparing equation 1 and equation 2.

F' = 9F/4

Hence the resulting mutual force change by a factor of 9/4

Answer: 9.1 × 10^-26 Joule

Explanation:

Since the collision is elastic. The kinetic energy will be conserved. That is, the sum of kinetic energy before collision will be the same as the sum of the

energy after collision.

Mass of an electron = 9.1 × 10^-31 kg

Given that the velocity of the moving electron = 400 m/s and the stationary electron now has a velocity = 200 m/s. 

K.E = 1/2mv^2

Add the two kinetic energies

1/2mV1^2 + 1/2mV2^2

1/2m( V1^2 + V2^2 )

Since they both have common mass

Substitute m and the two velocities

1/2 × 9.1×10^-31( 400^2 + 200^2)

4.55×10^-31 ( 160000 + 40000 )

4.55×10^-31 × 200000

K.E = 9.1 × 10^-26 Joule

Therefore, the initial kinetic energy of the moving electron is 9.1×10^-26 J

The complete question is missing, so i have attached the complete question.

Answer:

A) FBD is attached.

B) The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Explanation:

A) I've attached the image of the free body diagram.

B) The formula for the net force is given as;

F_net = mv²/r

We know that angular velocity;ω = v/r

Thus;

F_net = mω²r

Now, the minimum downward force is the weight and so;

mg = m(ω_min)²r

m will cancel out to give;

g = (ω_min)²r

(ω_min)² = g/r

ω_min = √(g/r)

The condition that must be satisfied is for ω_min = √(g/r)

C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).

Answer:

The total momentum is  [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]

Explanation:

The diagram illustration this  system is shown on the first uploaded image (From physics animation)

From the question we are told that

     The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]

      The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]

      The mass of the second object is  [tex]M_2 = 4 \ Dalton[/tex]

      The speed of the second object is  assumed to be  [tex]- v_2[/tex]

The total momentum of the system is the combined momentum of both object which is mathematically represented as

           [tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]

   substituting values

            [tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]

            [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

The ball reaches the maximum height of 54 feet

The question is about projectile motion,

the ball is shot at an angle α = 45°, and

the initial velocity u = 46 ft/s.

Under the projectile motion, the maximum height H is given by:

[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]

where, g = 9.8 m/s²

substituting the given values we get:

[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]

Hence, the maximum height is 54 feet.

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Answer:

F₁ = 100 N

Explanation:

The pressure must be equally transmitted from the piston to the narrow arm. Therefore,

P₁ = P₂

F₁/A₁ = F₂/A₂

F₁/F₂ = A₁/A₂

where,

F₁ = Force Required to be applied to piston = ?

F₂ = Force to push cork at narrow arm = 16 N

A₁ = Area of wider arm = πd₁²/4

A₂ = Area of narrow arm = πd₂²/4

Therefore,

F₁/16 N = (πd₁²/4)/(πd₂²/4)

F₁ = (16 N)(d₁²/d₂²)

but, it is given that the diameter of wider arm is 2.5 times the diameter of the narrow arm.

d₁ = 2.5 d₂

Therefore,

F₁ = (16 N)[(2.5 d₂)²/d₂²]

F₁ = (16 N)(6.25)

F₁ = 100 N

Answer:

The the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.

Explanation:

Given;

mass of bullet, m₁ = 4.2 g = 0.0042 kg

mass of hunter + gun = 72.5 kg

velocity of the bullet, u = 965 m/s

Momentum of the bullet when it was fired;

P = mv

P = 0.0042 x 965

P = 4.053 kg.m/s

Determine the recoil velocity of the hunter.

Total momentum = sum of the individual momenta

Total momentum = momentum of the bullet + momentum of the hunter

Apply the principle of conservation of momentum, sum of the momentum is equal to zero.

[tex]P_{hunter} + P_{bullet} = 0\\\\P_{hunter} = -P_{bullet}\\\\72.5v = -4.053\\\\v = \frac{-4.053}{72.5} \\\\v = - 0.056 \ m/s\\\\Thus, the \ recoil \ velocity \ of \ the \ hunter \ is \ 0.056 \ m/s, \ in \ opposite \ direction \ of \ the \ bullet.[/tex]

Therefore, the the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.

Answer:

C. explicit and implicit

Explanation:

E20

Answer:

Resultant of two vectors having opposite direction is the difference of the two displacements having the same direction as the larger vector.

Answer:

The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.

Explanation:

The period of oscillation is calculated as;

[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]

where;

L is the length of the pendulum bob

g is acceleration due to gravity

If we make L the subject of the formula in the equation above, we will have;

[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]

The length of the pendulum depends on acceleration due to gravity (g).

Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.

Answer:

T = 7.61*10^6 s

Explanation:

In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:

[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex]         (1)

T: period of Mercury

r: distance between Mercury and Sun

Ms: mass of the sun = 1.98*10^30 kg

G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2

You replace the values of all parameters in the equation (1):

[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]

The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

Answer:

a.     E = 1.02*10^-27 J

b.     E = 6.39*10^-9eV

Explanation:

a. In order to calculate the energy of the radio photon, you use the following formula:

[tex]E=hf[/tex]             (1)

h: Planck's constant = 6.626*10^-34 Js

f: frequency of the photon = 1545kHz = 1.545*10^6 Hz

Then, by replacing you obtain the energy of the photon:

[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]

b. In electron volts, the energy of the photon is:

[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from  different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

Answer:

   R _average = 1020 Ω

Explanation:

In your exercise, two circuits were assembled and you wanted to test the ohm's law equations with them, for those you surely wanted the resistance to be equal, the variations of these resistances ( R= 1 KΩ) are within the tolerance range of them, which for good resistances is 5% which gives a fluctuation of 50 Ω.

If more exact calculations are desired, the average of the resistance value can be used, which is the sum of the values ​​between the number of resistances

         R_average = 1 / n Σ [tex]R_{i}[/tex]

where R_{i} are the values ​​of each resistance and n the number of resistances.

        R_average = (1009 + 1031) / 2

        R _average = 1020 Ω

Answer:

a) P = 10.27 kW

b) Pmax = 10.65 kW

c) E = 5.47 MJ

Explanation:

Mass of the loaded car, m = 950 kg

Angle of inclination of the shaft, θ = 28°

Acceleration due to gravity, g = 9.8 m/s²

The speed of the car, v = 2.35 m/s

Change in time, t = 14.0 s

a) The power that must be provided by the winch motor when the car is moving at constant speed.

P = Fv

The force exerted by the motor, F = mg sinθ

P = mgv sinθ

P = 950 * 9.8 *2.35* sin28°

P = 10,271.3 W

P = 10.27 kW

b) Maximum power that the motor must provide:

[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]

c) Total energy transferred:

Length of the track, d = 1250 m

[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]

Answer:

Explanation:

We know that efficiency is the ratio of output power by input power. i.e. Efficiency describes the quality of machine or system how good it is.

Solution,

Energy input of system = 120 J

Efficiency = 80% = [tex] \frac{80}{100} = 0.8[/tex]

Now,

According to definition,

Efficiency = [tex] \frac{output}{input} [/tex]

Cross multiplication:

[tex]output \: = \: 0.8 \times 120[/tex]

Calculate the product

[tex]output \: = 96 \: joules[/tex]

Hope this helps...

Good luck on your assignment...

Answer:

t = 0.22 s

Explanation:

The rate of curvature can be defined as the ratio of the radius of curvature to the time taken to bend the wire to that radius of curvature. Therefore,

v = r/t

where,

v = rate of curvature

r = radius of wire

t = time taken

Here, in our case:

v = 9 units/s

r = 2 units

t = ?

Therefore,

9 units/s = 2 units/t

t = (2 units)/(9 units/s)

t = 0.22 s

Therefore, it takes 0.22 second to bend a straight wire into a circle of radius 2 units