Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 56 m long that spin at 13 rpm .A. At the tip of a blade, what is the speed?
B. At the tip of a blade, what is the centripetal acceleration?
C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .
D. What force is required to keep a 10 mg water droplet moving in this circular arc?
E. What is the ratio of this force to the weight of a droplet?

Answers

Answer 1
Answer:

A) The speed of the tip of the blade is 76.2 m/s

B) The centripetal acceleration of the tip of the blade is 103.7 m/s^2

D) The force required to keep the droplet moving in circular motion is 0.39 N

E) The ratio of the force to the weight of the droplet is 4.0

Explanation:

A)

We know that the blade of the turbine is rotating at an angular speed of

\omega = 13 rpm

First, we have to convert this angular speed into radians per second. Keeping in mind that

1 rev = 2 \pi

1 min = 60 s

We get

\omega = 13 rpm \cdot (2\pi rad/rev)/(60 s/min)=1.36 rad/s

The linear speed of a point on the blade is given by:

v=\omega r

where

\omega=1.36 rad/s is the angular speed

r is the distance of the point from the axis of rotation

For a point at the tip of the blade,

r = 56 m

Therefore, its speed is

v=(1.36)(56)=76.2 m/s

B)

The centripetal acceleration of a point in uniform circular motion is given by

a=(v^2)/(r)

where

v is the linear speed

r is the distance of the point from the axis of rotation

In this problem, for the tip of the blade we have:

v = 76. 2 m/s is the speed

r = 56 m is the distance from the axis of rotation

Substituting, we find the centripetal acceleration:

a=((76.2)^2)/(56)=103.7 m/s^2

D)

The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:

F=m(v^2)/(r)

where

m is the mass of the droplet

v is the linear speed

r is the distance from the centre of rotation

The data in this problem are

m = 10 mg = 0.010 kg is the mass of the droplet

v = 2.5 m/s is the linear speed

r = 16 cm = 0.16 m is the radius of the circular path

Substituting,

F=(0.010)(2.5^2)/(0.16)=0.39 N

E)

The weight of the droplet is given by

F_g = mg

where

m = 10 mg = 0.010 kg is the mass of the droplet

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

F_g = (0.010)(9.8)=0.098 N

The force that keeps the droplet in circular motion instead is

F = 0.39 N

Therefore, the ratio between the two forces is

(F)/(F_g)=(0.39)/(0.098)=4.0

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What are 4 ways individuals can influence the government?

Answers

Voting, running, speaking in front of government

A javelin is thrown in the air. Its height is given by ( ) 1 2 8 6 20 h x x x = − + + , where x is the horizontal distance in feet from the point at which the javelin is thrown. a. How high is the javelin when it was thrown? b. What is the maximum height of the javelin? c. How far from the thrower does the javelin strike the ground?'

Answers

The maximum height, the location on the ground and the initial vertical height of the javelin is required.

The initial height of the javelin is 6 feet.

The maximum height of the javelin is 326 feet.

The javelin strikes the ground at 160.75 feet.

The given equation is

h(x)=-(1)/(20)x^2+8x+6

where x is the horizontal distance

At x= we will get the initial vertical height.

h(0)=-(1)/(20)*0+8* 0+6\n\Rightarrow h(0)=6

Vertex of a parabola is given by

x=-(b)/(2a)=-(8)/(2* -(1)/(20))\n\Rightarrow x=80

h(80)=-(1)/(20)(80)^2+8* 80+6=326

At h(x)=0 the javelin will hit the ground

0=-(1)/(20)x^2+8x+6\n\Rightarrow -x^2+160x+120=0\n\Rightarrow x=(-160\pm √(160^2-4\left(-1\right)*120))/(2\left(-1\right))\n\Rightarrow x=-0.75,160.75

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This question is incomplete, the complete question is;

A javelin is thrown in the air. Its height is given by h(x) = -1/20x² + 8x + 6

where x is the horizontal distance in feet from the point at which the javelin is thrown.

a. How high is the javelin when it was thrown?

b. What is the maximum height of the javelin?

c. How far from the thrower does the javelin strike the ground?'

Answer:

a. height of the javelin when it was thrown is 6 ft

b. the maximum height of the javelin is 326 ft

c. distance from the thrower is 160.75 ft

Explanation:

a)

Given h(x) = -1/20x² + 8x + 6

we determine the height when x = 0

h(0) = -1/20(0)² + 8(0) + 6 = 6 ft

therefore height of the javelin when it was thrown is 6 ft

b)

to determine the maximum height of the javelin;

we find the vertex of the quadratic

so

h = - [ 8 / ( 2(-1/20) ) ] = 80

therefore

h(80) = -1/20(80)² + 8(80) + 6

= -320 + 640 + 6 = 326 ft

therefore the maximum height of the javelin is 326 ft

c)

Now the thrower is at the point ( 0,0 ) and the javelin comes down at another point ( x,0 )

this is possible by calculating h(x) = 0

⇒ -1/20x² + 8x + 6 = 0

⇒ x² - 160x - 120 = 0

⇒ x = [ -(-160) ± √( (-160)² - 4(1)(-120) ) ] /  [ 2(1) ]

x = [ 160 ± √(25600 + 480) ] / 2

so

[x = 160.75 ; x = -0.75 ]

distance cannot be Negative

therefore distance from the thrower is 160.75 ft

A 60kg woman on skates throws a 3.9kg ball with a velocity of37m.s west. What is the velocity of the woman?

Answers

Answer:

2.405 m/s

Explanation:

Given that,

Mass of a women, m₁ = 60 kg

Mass of a ball, m₂ = 3.9 kg

Velocity of the ball, v₂ = 37 m/s

We need to find the velocity of the woman. It is a concept based on the conservation of linear momentum. Let v₁ is the velocity of the woman. So,

m_1v_1=m_2v_2\n\nv_1=(m_2v_2)/(m_1)\n\nv_1=(3.9* 37)/(60)\n\nv_1=2.405\ m/s

So, the velocity of the woman is 2.405 m/s.

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________

Answers

The speed when all the fuel has been exhausted is 2415m/s

  • According to this question, the following information was given:

  1. Exhaust velocity of fuel, V(e)= 1500 m/s
  2. Initial speed of rocket, V₁ = 0 m/s
  3. Final speed of rocket, V₂ = ?
  4. Fuel weight = 80% of total weight

  • Using Tsiolkovsky rocket equation as follows:

∆V = V(e) ln(m1/m2)

  1. m1 = initial mass
  2. m2 = final mass without repellant

  • m2 = m1 - 80%m1

  • m2 = m1 - 0.8m1

  • m2 = 0.2m1

  • ∆V = V2 - V1

Hence;

  • V2 - 0 = 1500 × ln (m1/0.2m1)

  • V2 = 1500 ln(1/0.2)

  • V2 = 1500 × 1.609

  • V2 = 2415m/s.

  • Therefore, the speed when all the fuel has been exhausted is 2415m/s.

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Answer:

v_2 =2414\ m/s

Explanation:

given,

exhaust velocity of fuel(v_e) = 1500 m/s

initial speed of rocket,v₁ = 0 m/s

final speed of rocket, v₂ = ?

fuel weigh = 80 % of total weight

using  Tsiolkovsky rocket equation

\Delta v = v_e ln((m_1)/(m_2))

Δ v = v₂ - v₁

v_e is the exhaust speed

m₁ is the  initial total mass.

m₂ is the is the final total mass without propellant.

m₂  = m₁ - 0.8 m₁

m₂  = 0.2 m₁

v_2-v_1 = 1500* ln((m_1)/(0.2 m_1))

v_2 = 1500* ln((m_1)/(0.2 m_1))

v_2 =2414\ m/s

When all the fuel is exhausted speed of the fuel is equal to v_2 =2414\ m/s

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Answers

Answer:

v_(su) = 19.44 m/s

Explanation:

m_(su)=5.68x10^(29)kg\nm_(sa)=5.68x10^(26)kg

T=9.29x10^8\nr_(o)=1.43x10^(12)

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_(su)*r_(o)=(m_(sa)+m_(su))*r_(1)

solve to r1

r_1=(m_(sa)*r_(o))/(m_(sa)+m_(su))=(5.68x10^(26)*1.43x10^(12))/(5.68x10^(26)+5.68x10^(26))

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = (m_(sa)*2*pi*r_1^2)/(T)

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_(sun)=(m_(sa)*2*\pi *( 2r_(o)*r_1 -r_1^2))/(T)

Since

L_(su)= m_(su)*v_(su)*r_1

then

v_(su)=(m_(sa)*2*pi*(2r_(o)*r_(1)-r_(1)^2))/(T*m_(sa)*r_1)

v_(su) = 19.44 m/s

Final answer:

In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.

Explanation:

The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.

However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.

Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.

So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.

This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.

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At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

Answers

The molar latent enthalpy of boiling of iron at 3330 K is  ΔH = 342 * 10^3 J.

Explanation:

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

                      d ln p = (ΔH / RT^2) dt

                   (1/p) dp = (ΔH / RT^2) dt

                    dp / dt = p (ΔH / RT^2) = 3.72 * 10^-3

                  (p) (ΔH) / (8.31) (3330)^2 = 3.72 * 10^-3

                          ΔH = 342 * 10^3 J.