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3 years ago

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Consider two wind turbines, each located in sites with the same wind speed, and each with the same efficiency. Turbine 1 has a r

adius of 25 meters. Turbine 2 has a radius of 50 meters. Turbine 2 generates ___ times as much power as turbine 1. (Fill in the blank.)

1 answer:

Alex787 [66]3 years ago

7 0

Answer:

shut up nerd

Exption:

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Scientists can use spectral analysis of stars to determine which of the following?

stellarik [79]

The answer would be A)

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3 years ago

Read 2 more answers

If you drop a rock off a cliff and it hits the ground 3 seconds later how tall is that cliff

skelet666 [1.2K]

Answer:

Basic kinematics, negating drag and assuming ideal conditions, we use the equation:

d=vi*t+1/2*a*t^2

Since vi is 0 (we know this because you’re dropping it, not throwing it)…

…and the only acceleration acting on it is gravity, a=9.8 m/s^2…

…we get

d=1/2(9.8)(5)^2

Explanation:

Some quick mental math tells us that this is about 125 m.

Plugging it in, we find it to be 122.5 m.

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3 years ago

zimovet [89]

Answer:

hindi ko gets

Explanation:

ang gulo naman nyan

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2 years ago

A 150 g ball and a 240 g ball are connected by a 33-cm-long, massless, rigid rod. The balls rotate about their center of mass at

stira [4]

Answer:

v= 3.18 m/s

Explanation:

Given that

m= 150 g = 0.15 kg

M= 240 g = 0.24 kg

Angular speed ,ω = 150 rpm

The speed in rad/s

$\omega =\dfrac{2\pi N}{60}$

$\omega =\dfrac{2\pi \times 150}{60}$

ω = 15.7 rad/s

The distance of center of mass from 150 g

$r=\dfrac{150\times 0+240\times 33}{150+240}\ cm$

r= 20.30 cm

The speed of the mass 150 g

v= ω r

v= 20.30 x 15.7 cm/s

v= 318.71 cm/s

v= 3.18 m/s

6 0

3 years ago

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 700 kg. The satellite is initial

MaRussiya [10]

Answer:

your friend did a work of 1717.34 J

Explanation:

The sum of the work made by your friend and you is equal the change in the kinetic energy, so:

$W_a+W_b=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Where $W_a$ is the work that you did, $W_b$ is the work that your friend did, m is the mass of the satellite and $v_f$ and $v_i$ are the final and initial speeds

Additionally the work that you did is equal to:

$W_a=F.s$

Where F and s are vectors. So, the vector s that said the displacement is calculated as:

s = <2.3, 2.6, -11.8> - <3.3, -1.6, 2.0>

s = <-1 , 4.2 , -13.8>

Therefore, $W_a$ is:

$W_a=.$

$W_a=(-400*-1)+(500*4.2)+(250*-13.8)\\W_a=400+2100-3450\\W_a=-950$

Then, replacing the values on initial equation and solving for $W_b$ , we get:

$-950+W_b=\frac{1}{2}(700)(6.18)^{2}-\frac{1}{2}(700)(6)^{2}\\-950+W_b=767.24\\W_b=767.24+950\\W_b=1717.34J$

Finally, your friend did a work of 1717.34 J

3 0

2 years ago