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3 years ago

10

A wind turbine is rotating counterclockwise at 0.626 rev/s and slows to a stop in 12.9 s. Its blades are 17.9 m in length. What

is the centripetal acceleration of the tip of the blades at t=0~\text{s}t=0 s?

1 answer:

iogann1982 [59]3 years ago

5 0

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

$\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s$

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

$a=\omega^2 r$

At the instant t = 0,

$\omega=3.93 rad/s$

So the centripetal acceleration of the tip of the blades is

$a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2$

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Jet001 [13]

Answer:

The cost of the buying the shirts is #9680

Explanation:

let the cost of buying shirt = C

let the number of shirt bought = N

The following equation can be generated based on the statement above;

C = k + Nb

When the cost, C = #240, the number of shirt = 5

240 = k + 5b ------ equation (1)

where;

k and b are constants

When the cost, C = #400, the number of shirt = 10

400 = k + 10b ------ equation (2)

From equation (1), make k the subject of the formula;

k = 240 - 5b ---- equation (3)

Substitute in the value of k into equation (2)

400 = k + 10b

400 = (240 - 5b) + 10b

400 = 240 - 5b + 10b

400 - 240 = -5b + 10b

160 = 5b

b = 160 / 5

b = 32

From equation (3), calculate k

k = 240 - 5b

k = 240 -5(32)

k = 240 - 160

k = 80

When the number of shirts bought = 300, the cost of the buying the shirts =

C = k + Nb

C = 80 +32N

Where;

N is the number of shirts

C = 80 + 32(300)

C = 80 + 9600

C = #9680

Therefore, the cost of the buying the shirts is #9680

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3 years ago

One kg of air contained in a piston-cylinder assembly undergoes a process from an initial state whereT1=300K,v1=0.8m3/kg, to a f

lutik1710 [3]

Answer:

1. Yes, it can occur adiabatically.

2. The work required is: 86.4kJ

Explanation:

1. The internal energy of a gas is just function of its temperature, and the temperature changes between the states, so, the internal energy must change, but how could it be possible without heat transfer? This process may occur adiabatically due to the energy balance:

$U_{2}-U_{1}=W$

This balance tell us that the internal energy changes may occur due to work that, in this case, si done over the system.

2. An internal energy change of a gas may be calculated as:

$du=C_{v}dT$

Assuming $C_{v}$ constant,

$U_{2}-U_{1}=W=m*C_{v}(T_{2}-T_{1})$

$W=0.72*1*(420-300)=86.4kJ$

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4 years ago

Which of the following items conducts thermal energy best?

kompoz [17]

Its b i belive

because it the only thing i saw on the list that conduts

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3 years ago

Read 2 more answers

A ball has a diameter of 3.79 cm and average density of 0.0838 g/cm3.

suter [353]

Answer: 0.258 N

Explanation:

As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.

So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:

F = Fb – Fg

Fb= δH20* 4/3*π*(d/2)³ * g

Fg = δb* 4/3*π*(d/2)³ *g

F= (δH20- δb) * 4/3*π*(d/2)³*g

Replacing by the values of the densities, and the ball diameter, we finally get:

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4 years ago

A rock is dropped from the top of a vertical cliff and takes 3.00 s to reach the ground below the cliff. A second rock is thrown

Trava [24]

Answer:

D) 12.3 m/s downward

Explanation:

We use the next free fall equation

$h=v_{0}t+\frac{1}{2}gt^2$

where $h$ is the height of the cliff, $v_{0}$ the initial velocity, $g$ the acceleration of gravity ( $g=9.81m/s^2$ ) and $t$ is time.

For the fist rock $v_{0}=0$ since the rock was dropped, and $t=3s$ , so we have:

$h=(0m/s)(3s)+\frac{1}{2}(9.81m/s^2)(3s)^2$

simplifying

$h=44.145m$

the height of the cliff is 44.145m

Now, about the second rock we know that is the same height and now the time es $t=2s$ , and we need to find $v_{0}$

From the fist equation

$h=v_{0}t+\frac{1}{2}gt^2$

we clear for $v_{0}$

$v_{0}t=h-\frac{1}{2} gt^2\\v_{0}=\frac{h}{t} -\frac{1}{2} gt$

and substitute known values

$v_{0}=\frac{44.145m}{2s} -\frac{1}{2} (9.81m/s^2)(2s)$

$v_{0}=22.07m/s -9.81m/s$

$v_{0}=12.26m/s$ wich rounds up to $v_{0}=12.3m/s$

the direction is downward because the rock is thrown so that it falls through the cliff.

7 0

3 years ago