At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 90-m-diameter (D) blades at that location. Take the air density to be 1.25 kg/m3. The mechanical energy of air per unit mass is .05 Numeric ResponseEdit Unavailable. .05 correct.kJ/kg. The power generation potential of the wind turbine is

Answer:

Let [tex]A[/tex], [tex]B[/tex],  and [tex]k[/tex] denote three constants (with the requirement that [tex]k > 0[/tex].) The following assumes that the mass of this object is [tex]m[/tex]. Assume that [tex]x(t)[/tex] denotes the position of the object at time [tex]t[/tex].

(a) [tex]\displaystyle x^\prime(t) = -\sqrt{\frac{k}{m}}\, A\, \sin\left(t\, \sqrt{\frac{k}{m}}\right) + \sqrt{\frac{k}{m}}\, B\, \cos\left(t\, \sqrt{\frac{k}{m}}\right)[/tex].

(b) [tex]\displaystyle x^{\prime\prime}(t) = -\left(\frac{k}{m}\right)\, A\, \cos\left(t\, \sqrt{\frac{k}{m}}\right) - \left(\frac{k}{m}\right)^{2}\, B\, \sin\left(t\, \sqrt{\frac{k}{m}}\right)[/tex]

Explanation:

The differential equation for a simple harmonic motion might take the following form:

[tex]\displaystyle \frac{d^{2} x}{d t^{2}} = -\frac{k}{m}\, x[/tex].

The minus sign on the right-hand side highlights the fact that the displacement and acceleration of the object should be in opposite directions.

Notice how this equation is in the form of a homogeneous second-order ODE:

[tex]x^{\prime\prime}(t) + \underbrace{P(t)}_{0}\, x(t) + \underbrace{Q(x)}_{\sqrt{k / m}} = 0[/tex]

Let [tex]r[/tex] be a constant. One possible solution to this homogeneous second-order ODE would be in the form [tex]x(t) = e^{r\, t}[/tex], such that [tex]x^{\prime}(t) = r\, e^{r\, t}[/tex] whereas [tex]x^{\prime\prime}(t) = r^{2}\, e^{r\, t}[/tex].

Substitute into the original ODE to obtain:

[tex]\displaystyle \underbrace{r^{2}\, e^{r t}}_{x^{\prime\prime}(t)} + \frac{k}{m}\, \underbrace{e^{r t}}_{x(t)} = 0[/tex].

Rearrange the equation and solve for [tex]r[/tex].

[tex]\displaystyle e^{r t}\, \left(r^{2} + \frac{k}{m}\right) = 0[/tex]

Notice that [tex]e^{r\, t} > 0[/tex]. Hence, it must be true that [tex]\displaystyle r^{2} + \frac{k}{m} = 0[/tex]. Solve for [tex]r[/tex] given that [tex]k > 0[/tex]:

[tex]\displaystyle r_{1, 2} = \pm i\sqrt{\frac{k}{m}}[/tex], where [tex]i[/tex] is the imaginary unit.

The two particular solutions for the ODE would be:

[tex]x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}[/tex] and [tex]x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}[/tex].

Apply Euler's Formula to rewrite both solutions in terms of trigonometric functions:

[tex]\displaystyle x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}= \sqrt{\frac{k}{m}} \left(\cos\left( \sqrt{\frac{k}{m}} t\right) + + i\, \sin\left( \sqrt{\frac{k}{m}} t\right)\right)[/tex].

[tex]\displaystyle x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}= -\sqrt{\frac{k}{m}} \left(\cos\left( -\sqrt{\frac{k}{m}} t\right) + i\, \sin\left( -\sqrt{\frac{k}{m}} t\right)\right)[/tex].

The general solution would be in the form:

[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \left(i\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)\right)[/tex],

Where [tex]C_1[/tex] and [tex]C_2[/tex] are constants (not necessarily real numbers.)

Since position is supposed to assume a real value for any real [tex]t[/tex], set [tex]B[/tex] to a multiple of [tex]i[/tex] such that the general solution is real-valued:

[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)[/tex].

Differentiate to obtain general expressions for velocity (first derivative) and acceleration (second derivative.)

some positions are quarterback, offence, defense, linebacker, and wider receiver hoped it helped :)

Answer:

37.33m

Explanation:

Using the equation of motion

S = ut + 1/2gt^2

Time t = 2.76secs

g = 9.8m/s^2

S = 0 + 1/2(9.8)(2.76)^2

S = 4.9*7.6176

S = 37.33

Hence the window is 37.33m above the ground

Answer:

v = 5.7554 m/s

Explanation:

First of all we need to know if the angle of the vine is measured in the horizontal or vertical.

To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:

h₀ = h sinα

h₀ = 5.32 sin43° = 3.6282 m

As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:

Δh = 5.32 - 3.63 = 1.69 m

In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:

Ep = Ek

From here, we can calculate the speed of the monkey.

Ep = mgΔH

Ek = 1/2 mv²

The potential energy is:

Ep = 16.9 * 9.8 * 1.69 = 279.9

Now with the kinetic energy:

1/2 * (16.9) * v² = 279.9

v² = (279.9) * 2 / 16.9

v² = 33.12

v = √33.12

Hope this helps

Answer: Change occurs when forces in one direction exceed forces in opposing direction

Explanation:

Force field analysis is the basic tool typically used in root cause analysis which enable an individual to take certain action after the identification of the root cause.

According to force-field analysis, there is a change when the forces in one direction exceed the forces in opposing direction.

Answer:

a= 17.69 m/s^2

Explanation:

Step one:

given data

A car accelerates uniformly from rest to 23 m/s

u= 0m/s

v= 23m/s

distance= 30m

Step two:

We know that

acceleration= velocity/time

also,

velocity= distance/time

23= 30/t

t= 30/23

t= 1.30 seconds

hence

acceleration= 23/1.30

accelaration= 17.69 m/s^2

Answer:

the speed of electron is 4.42 x 10⁶ m/s

the speed of proton is 2406.7 m/s

Explanation:

Given;

electric field strength, E = 478 N/C

charge of the particles, Q = 1.6 x 10⁻¹⁹ C

mass of proton, Mp = 1.673 x 10⁻²⁷ kg

mass of electron Me = 9.11 x 10⁻³¹ kg

time of motion, t = 54.2 ns = 54.2 x 10⁻⁹ s

The magnitude of charge experienced by the particles is calculated as;

F = EQ

F = 478 x 1.6 x 10⁻¹⁹

F = 7.648 x 10⁻¹⁷ N

The speed of the particles is calculated as;

[tex]F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{9.11\times 10^{-31}} \\\\v_e = 4.42 \ \times \ 10^6 \ m/s[/tex]

[tex]v_p = \frac{Ft}{m_p} \\\\v_p = \frac{(7.684 \times 10^{-17})(52.4\times 10^{-9})}{1.673\times 10^{-27}} \\\\v_p = 2406.7 \ m/s[/tex]

Answer: Charge 3 is located on the x-axis a distance of 0.67 cm from charge 1 and 1.33 cm from charge 2.

Explanation: Electrostatic Force is the force of repulsion or attraction between two charged particles. It's directly proportional to the charge of the particles and inversely proportional to the distance between them:

[tex]F=k\frac{|q||Q|}{r^{2}}[/tex]

k is an electrostatic constant

For the system of 3 particles, suppose distance from 1 to 3 is x meters, so, distance from 2 to 3 is (0.02-x) meters.

Force will be

[tex]F_{13}=F_{23}[/tex]

[tex]k\frac{q_{1}q_{3}}{r_{13}^{2}} =k\frac{q_{2}q_{3}}{r_{23}^{2}}[/tex]

[tex]\frac{q_{1}}{r_{13}^{2}} =\frac{q_{2}}{r^{2}_{{23}}}[/tex]

Substituting:

[tex]\frac{2.10^{-6}}{x^{2}} =\frac{8.10^{-6}}{(0.02-x)^{2}}[/tex]

[tex]8.10^{-6}x^{2}=2.10^{-6}(0.0004-0.04x+x^{2})[/tex]

[tex]4x^{2}=x^{2}-0.04x+0.0004[/tex]

[tex]3x^{2}+0.04x-0.0004=0[/tex]

Solving quadratic equation using Bhaskara:

[tex]x_{1}=\frac{-0.04+\sqrt{(0.04)^{2}+0.048} }{6}[/tex]

[tex]x_{2}=\frac{-0.04-\sqrt{(0.04)^{2}+0.048} }{6}[/tex]

x₂ will give a negative value and since distance can't be negative, use x₁

[tex]x_{1}=\frac{-0.04+\sqrt{0.0064} }{6}[/tex]

x₁ = 0.0067 m

The position of charge 3 is 0.67 cm from charge 1 and 1.33 cm from charge 2.

Answer:

F • t= m•∆v

Explanation:

the impulse experienced by the object equals the change in momentum of the object. in equation form: F • t= m•∆v

Answer:

The  velocity of the waves in the string is 30 m/s.

Explanation:

Given;

frequency of the wave, F = 60.0 Hz

number of loops, n = 5

length of the string, L = 1.25 m

1 loop, L  = λ/2

For 5 loops;

[tex]L = \frac{5\lambda}{2}[/tex]

where;

λ is the wavelength

[tex]L = \frac{5 \lambda}{2} \\\\5 \lambda = 2L\\\\\lambda = \frac{2L}{5} \\\\\lambda = \frac{2 \ \times \ 1.25}{5} \\\\\lambda = 0.5 \ m[/tex]

The velocity of the waves in the string is calculated as;

V = Fλ

V = (60)(0.5)

V = 30 m/s

Therefore, the  velocity of the waves in the string is 30 m/s.

Answer:

Collection. or Precipitation.

Explanation:

b or d

The part of the water cycle when ocean water may end up in the river is to be considered as the condensation.

It is the change with regard to the state of matter from the gas phase into the liquid phase. Also, it should be considered as the reverse of vaporization.

So based on this, we can say that The part of the water cycle when ocean water may end up in the river is to be considered as the condensation.

Learn more about water here: https://brainly.com/question/26094433

Answer:

Mass of player 2 is 55.044kg

Explanation:

Step one:

given data

Weight of player 1= 900 N

mass= weight/acceleration due to gravity

mass of player= 900/9.81

m1= 91.74kg

v1= 8m/s

mass of player 2

m2= ?

v2= 0m/s

Step two:

from the problem description, after impact, they both moved with a common velocity of V= 5m/s, hence the collision is an inelastic collision

the expression for inelastic collision is

m1v1+m2v2=(m1+m2)V

substitute

91.74*8+m2*0=(91.74+m2)5

733.92=458.7+5m2

275.22=5m2

divide both sides by 5

m2=275.22/5

m2=55.044kg

The weight of player two is

=55.044*9.81

=539.98N

Answer:

If there is a large difference between the concentration of one fluid and the concentration of the other, then the particles will diffuse faster. In the experiment, the dye particles in the food coloring are very concentrated.

Explanation:

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

[tex]Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd[/tex]

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

[tex]\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}[/tex]

d = 10.2 m

Answer:

21977.56kg⋅m/s

Explanation:

Step one:

Given data

Weight of car= 9800N

The mass of the car is

m= w/g

m= 9800/9.81

m=998.98kg

Velocity of car= 22 m/s

time of impact= 0.5 seconds

Required

The impulse of the force F

Step two:

Ft= mv

Ft= 998.98*22

Ft=21977.56kg⋅m/s

The impulse is 21977.56kg⋅m/s

Answer: I think this is the answer, In a collision, there is a force on both objects that causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Explanation: I had a question similar to this, Hope this helps!

If we compare the energy of visible light to the energy of X-rays, we find that X-rays have a much higher frequency. Usually, electromagnetic radiation with higher frequency (energy) have a higher degree of penetration than those with low frequency.

Since Saturn's volume is equivalent to 763.59 earths and Mercury is only 0.055 of one earth, then Saturn's volume is over 10,000 times larger than Mercury's.

Answer: TRUE

The work done by the given force should be -45 joules

Since force acts in the negative y direction

So,

F = -15j

here the displacement vector is d = 3i+3j-1k

Now

work done is W = dot product of force and displacement

                        = [-15j ] . [ 3i+3j-1k]

                        =-45 joules

Therefore, The work done by the given force should be -45 joules

Learn more about force here:

https://brainly.com/question/93252

Answer:

50.97 m

Explanation:

m = Mass of truck

[tex]\mu_s[/tex] = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = [tex]\dfrac{72}{3.6}=20\ \text{m/s}[/tex]

s = Displacement

Force applied

[tex]F=ma[/tex]

Frictional force

[tex]f=\mu_s mg[/tex]

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

[tex]ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2[/tex]

Since the obect will be decelerating the acceleration will be [tex]-3.924\ \text{m/s}^2[/tex]

From the kinematic equations we have

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}[/tex]

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

Answer:

.

Explanation:

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 [tex]\pi[/tex] [tex]r^{2}[/tex]  = [tex]\frac{Q1}{E_{0} }[/tex]

So,

Rearranging the above equation to get Electric field, we will get:

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex]

Multiply and divide by [tex]r1^{2}[/tex]

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] x [tex]\frac{r1^{2} }{r1^{2} }[/tex]

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

c) r > r2 :

Electric Field = ?

E x 4 [tex]\pi[/tex] [tex]r^{2}[/tex]  = [tex]\frac{Q1 + Q2}{E_{0} }[/tex]

Rearranging the above equation for E:

E = [tex]\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }[/tex]

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] + [tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex]

As we know from above, that:

[tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] =  (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

Then, Similarly,

[tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex] = (σ2 x [tex]r2^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

So,

E = [tex]\frac{Q1}{E_{0} . 4 \pi. r^{2} }[/tex] + [tex]\frac{Q2}{E_{0} . 4 \pi. r^{2} }[/tex]

Replacing the above equations to get E:

E = (σ1 x [tex]r1^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex]) + (σ2 x [tex]r2^{2}[/tex]) /([tex]E_{0}[/tex] x [tex]r^{2}[/tex])

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x [tex]r1^{2}[/tex] = - σ2 x [tex]r2^{2}[/tex]

Answer:

5

Explanation:

Answer:

the frequency of the beats is 8.7687 kHz

Explanation:  

Given the data in the question;

The frequency for stationary source and moving observer is;

f' = f( 1 +( v_observer/v_sound))

we know that, speed of sound in dry air =  343 m/s

so we substitute

f' = 41.2 kHz( 1 + (33.0 m/s / 343 m/s) ) = 41.2kHz( 1 + 0.0962) = 41.2kHz(1.0962)

f' = 45.1634 kHz

Now the frequency for stationary observer and moving source with frequency f' will be

f" = f'( 1 / (1 - ( v_observer/v_sound)))

45.1634 kHz( 343 / 343 - 33)

we substitute

f" = 45.1634 kHz( 1 / (1 - (33.0 m/s / 343 m/s)))

f" = 45.1634 kHz( 1 / (1 - 0.0962))

f" = 45.1634 kHz( 1 / 0.9038 )

f" = 45.1634 kHz( 1.1064 )

f" = 49.9687 kHz

Now the beat frequency will be;

f_beat = f' - f

we substitute

f_beat = 49.9687 kHz - 41.2 kHz

f_beat = 8.7687 kHz

Therefore, the frequency of the beats is 8.7687 kHz

Answer:

44.5s ; 22.64 m/s

Explanation:

The motion of the truck could be separated into 3 different phases :

First :

Time of motion :

Initial Velocity, u = 0 ; final velocity, v = 29 m/s

Acceleration, a = 2 m/s²

Recall: acceleration = change in velocity / time

Time = change in velocity / acceleration

Time = (29 - 0) / 2 = 14.5 second

Distance traveled = ((29 + 0) /2) * 14.5 = 210.25 m

Second :

Time = 25 seconds at constant speed

29 m/s for 25 seconds

v*t = 29 * 25 = 725 m

Third:

5 seconds before coming to rest

((29 + 0) /2) * 5

14.5 * 5 = 72.5 m

A.)

Length of journey = (14.5+ 25 + 5) = 44.5 seconds

B.)

Average velocity = total distance / total time taken

Average velocity = (210.25 + 725 + 72.5) / 44.5

= 1007.75 / 44.5

= 22.646067

= 22.64 m/s

Answer:

392 Nm or J

Explanation:

Work is equal to force times distance. The force required is the mass times acceleration - in this case F = 5kg x 9.81 ( gravity) = 49N. So work is 49N x 8m = 392 Nm or Joules.

Answer:

10.44 liters of gasoline.

Explanation:

First, we need to convert the units of the car from mi/gal to km/L as follows:

[tex] mileage = 32.0 \frac{mi}{gal}*\frac{1.609 km}{1 mi}*\frac{1 gal}{3.785 L} = 13.60 km/L [/tex]

That means that for every liter of gasoline the car travels 13.60 kilometers.

So, to complete a 142-km trip in Europe the volume of gasoline needed is:

[tex] V = \frac{1 L}{13.60 km}*142 km = 10.44 L [/tex]

Therefore, to complete a 142-km trip in Europe we need to buy 10.44 liters of gasoline.

I hope it helps you!                

Answer:

117.8 m/s

Explanation:

Given that:

String length, L = 0.28

pipe length, L' = 0.82

Speed of sound in air, v = 345 m/s

n = 3 (3rd harmonic)

Frequency, f of 3rd harmonic ;

f = (v*n) / 2L - - - - (1)

for the pipe: ; 3rd harmonic

f = (v*n) / 2L' - - - (2)

Equating (1) and (2)

(v*n) / 2L = (v*n) / 2L'

2L' * v * n = v* n * 2L

v = vL / L'

v = (345 * 0.28) / 0.82

v = 96.6 / 0.82

v = 117.80487

v = 117.8 m/s