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3 years ago

Why are satellites placed into orbit at least 150 km above Earth’s surface?

Physics

2 answers:

eduard3 years ago

8 0

The correct answer is to avoid air resistance

Satellites are moving in the orbit because of the initial force that was applied through the rockets. Now for the satellites to remain in orbit there should not be any friction. As air resistance is also friction that is provided by air, the satellites are 150KM above the surface to avoid air resistance

kogti [31]3 years ago

7 0

Satellites are placed into orbit at least 150 km
above Earth’s surface to avoid air resistance.

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Fragments of molecules that have at least one electron available for bonding are called

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They are called radicals.................

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3 years ago

Which one i need help

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Answer:

Explanation:

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3 years ago

A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate

PilotLPTM [1.2K]

Answer:

$T(1)=21$

Explanation:

The equation of the position in kinematics is given:

$x(t)=x_{0}+v_{0}t+0.5at^{2}$

x(0) is the initial position, in this it is 0
v(0) is the initial velocity (20 m/s)
a is the acceleration (2 m/s²)

So the equation will be:

$x(t)=20t+0.5*2*t^{2}$

$x(t)=20t+t^{2}$

Now, the Taylor polynomial equation is:

$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

$f(t)=x(t)=20t+t^{2}$

$f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t$

$f''(t)=\frac{dv(t)}{dt}=a(t)=2$

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

$f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}$

$f(0)=x(0)=0$

$f'(0)=v(0)=20$

$f''(0)=a(0)=2$

$T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}$

$T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}$

$T(t)=20t+t^{2}$

Let's put t=1 so find the how far the car moves in the next second:

$T(1)=20*1+1^{2}$

$T(1)=21$

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!

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3 years ago

A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th

Elis [28]

Answer:

The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

$y=y_0+V_0t-gt^2/2$

We are only interested in the vertical motion, so that equation is all what you need.

1. Water ballon is thrown horizontally at sped 2.00 m/s.

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

$y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s$

2. Water ballon thrown straight down at 2.00 m/s

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

$0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s$

To solve the equation you can use the quadratic formula.

$t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92$

You get two times. One of the times is negative, thus it does not have physical meaning.

3. Conclusion:

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

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Answer:

Remember: in physics and music, frequency equates to pitch, and amplitude equates to loudness. Also, remember frequency is inversely proportional to wavelength- remember this!

Thus, (B) should be the correct answer, because, think about it, if a note is higher, than the wavelength should be shorter so the length should also be shorter, but in the case of B, if the cable is longer or thicker then the wavelength gets longer. Therefore, going from low to high, the string should be getting shorter, due to the fact that as wavelength gets shorter, frequency gets higher.

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3 years ago