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ryzh [129]
3 years ago
14

How does a windmill work

Physics
1 answer:
saveliy_v [14]3 years ago
7 0
The windmills used mostly for pollution problems and farming and in a windmill there are 3 barrels that spin, but the barrels are tilted a little bit so when i spins it makes the wind go forward
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What is the frequency of a wave if 10 waves go past in 1 second ?
KIM [24]

Answer:

f = 10 Hz

Explanation:

We need to find the frequency of a wave if 10 waves go past in 1 second. Total number of waves per unit time is called frequency. So,

f=\dfrac{10}{1}=10\ Hz

So, the frequency of a wave is 10 Hz.

5 0
3 years ago
Please help !!! HELP
kozerog [31]

Answer:

The answer is C

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3 years ago
A satellite in earth orbit has a mass of 99 kg and is at an altitude of 2.02 106 m. (assume that u = 0 as r â â.) (a) what is th
Ierofanga [76]

(a)          What is the potential energy: PE = -G * M * m/r

 

Where: M is the mass of the earth which is 5.98 * 10^24 kg.

m is the mass of the satellite.

r is the space from the center of the earth to the satellite

 

To conclude this distance add the radius of the earth to the altitude. Radius of the earth is 6.38 * 10^6 meters.

 

 

r = 6.38 * 10^6 + 2.02 * 10^6 = 8.38 * 10^6

PE = 6.67 * 10^-11 * 5.98 * 10^24 * 99/8.38 * 10^6 = 4.71240095 * 10^9 J

 

 

(b) magnitude of the gravitational force exerted by the Earth

 

Fg = G * M * m/r^2

Fg = 6.67 * 10^-11 * 5.98 * 10^24 * 99/(8.38 * 10^6)^2 = 562.3078873 N

 

 

(c) There are no other forces that the satellite exert on the Earth. So therefore, it is 0.

8 0
4 years ago
The product side of a chemical reaction is shown. → 7Ti2(SO4)3
Alex_Xolod [135]
The answer is the fourth choice because there are 7 represents in a coefficient.

7 0
3 years ago
Read 2 more answers
A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

6 0
3 years ago
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