There are strong winds on these trees. What makes the wind stronger? Why are they blowing from different directions?

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ

LenaWriter [7]

The orbital period of the satellite[T] is given as $2*10^{6} S$ .

The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V] written as -

V $=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

$V=\frac{2*3.14*10^{12} }{2*10^{6} }$

$= 7.85*10^{6} m/s$

There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

⇒ $GM=gR^{2}$ where g is the acceleration due to gravity of that central object.

Hence $V=\sqrt{\frac{gR^{2} }{R} }$

⇒ $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.

7 0

3 years ago

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Ce 1. A satellite does not need any energy to revolue around the planet. Why?

olasank [31]

Hope this helped! i’m not sure if it’s right. it honestly bored so i’m answering questions.

3 0

3 years ago

A sumo wrestler is 330 pounds. How much force does the earth exert in him? Define Mass Define Weight

Oksana_A [137]

Answer:

a large matters of body with no shape is mass

5 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

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Pls help me! I’ll mark branniest thank u!

Harrizon [31]

Answer:

D

Explanation:

Can you please mark brainiest

and I think it's d because you were at the same sowed as the car but as it turn you moved the other way because of the pull of friction and your weight

hope it helps

4 0

3 years ago

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