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1 year ago

What speed should a satellite of mass 4,900 kg moving around

1 answer:

8 0

Based on the calculations, the speed required for this satellite to stay in orbit is equal to 1.8 × 10³ m/s.

<u>Given the following data:</u>

Gravitational constant = 6.67 × 10⁻¹¹ m/kg²

Mass of Moon = 7.36 × 10²² kg

Distance, r = 4.2 × 10⁶ m.

<h3>How to determine the speed of this satellite?</h3>

In order to determine the speed of this satellite to stay in orbit, the centripetal force acting on it must be sufficient to change its direction.

This ultimately implies that, the centripetal force must be equal to the gravitational force as shown below:

Fc = Fg

mv²/r = GmM/r²

<u>Where:</u>

m is the mass of the satellite.

M is mass of the Moon.

Making v the subject of formula, we have;

v = √(GM/r)

Substituting the given parameters into the formula, we have;

v = √(6.67 × 10⁻¹¹ × 7.36 × 10²²/4.2 × 10⁶)

v = √(1,168,838.095)

v = 1,081.13 m/s.

Speed, v = 1.8 × 10³ m/s.

Read more on speed here: brainly.com/question/20162935

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

5 0

2 years ago

What is an independent variable

Iteru [2.4K]

Answer:

An independent variable is a variable that is manipulated to determine the value of a dependent variable. The dependent variable is what is being measured in an experiment or evaluated in a mathematical equation and the independent variables are the inputs to that measurement.

Explanation:

8 0

3 years ago

Read 2 more answers

Describe how impacts between objects in the solar system have helped create notable features in the solar system such as craters

konstantin123 [22]

When 'The big bang' happened lots of large pieces of molten rock was flying around the solar system. As the rocks crashed together they got bigger and as the got bigger they attracted more rocks. Some scientists think that a large piece of molten rock hit the still developing Earth and created the Moon. This impact also caused the Earths angled spin. The Moon got trapped in Earth's orbit and has stayed ever since. Small astroids have hit the Moon causing craters. The Earth doesn't get hit as much because of our thicker atmosphere. Hope this helps!

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2 years ago

A long cylindrical insulator has a uniform charge density of 1.54 µc/m3 and a radius of 6cm what is the electric field inside th

Lubov Fominskaja [6]

Plug given values in SI units to get electric field.

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2 years ago

The minimum breaking distance d in feet for a typical vehicle can be modeled with the equation d(v) = 0.045v2, where v is the ve

xxTIMURxx [149]

<span>To find the kind of transformation that describes this change from d(v) = 0.045v2 to d(v) = 0.039v2, find the relation between the two functions:

0.045/.039 = 45/39 = 15/13

The you have to multiply the first function times 13/15 to transform it to the second function.

When you multiply by a factor less than one you are compressiong the function vertically (if you multiply by a factor greater than 1 you are stretching vertically).

On the other hand, that the distance to stop the minimum braking distance will be smaller with the second function.

Then, the answer is that the transformation is a vertical compression by a factor of 13/15 and the braking distance will be less with optimum new tires than with tires having more wear.
</span>

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3 years ago