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Murrr4er [49]
2 years ago
8

Look at the pic please

Physics
1 answer:
SCORPION-xisa [38]2 years ago
7 0

Answer:

<h2><em><u>y</u></em><em><u>o</u></em><em><u>u</u></em><em><u>r</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>B</u></em><em><u>)</u></em><em><u> </u></em><em><u>i</u></em><em><u> </u></em><em><u>h</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>h</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u>s</u></em><em><u> </u></em><em><u>y</u></em><em><u>o</u></em><em><u>u</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>h</u></em><em><u>a</u></em><em><u>v</u></em><em><u>e</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>g</u></em><em><u>o</u></em><em><u>o</u></em><em><u>d</u></em><em><u> </u></em><em><u>d</u></em><em><u>a</u></em><em><u>y</u></em><em><u>.</u></em></h2>
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Calculate the velocity of a 1650kg satellite that is in a circular orbit of 4.2 x 10^6m above the surface of a planet which has
Anastasy [175]
-- We're going to be talking about the satellite's speed. 
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.

-- The mass of the satellite makes no difference.

Since the planet's radius is  3.95 x 10⁵m  and the satellite is
orbiting  4.2 x 10⁶m  above the surface, the radius of the
orbital path itself is

                               (3.95 x 10⁵m) + (4.2 x 10⁶m)

                     =        (3.95 x 10⁵m) + (42 x 10⁵m)

                     =           45.95 x 10⁵ m

The circumference of the orbit is  (2 π R) =  91.9 π x 10⁵ m.

The bird completes a revolution every 2.0 hours,
so its speed in orbit is

                                     (91.9 π x 10⁵ m) / 2 hr

                        =        45.95 π x 10⁵  m/hr  x  (1 hr / 3,600 sec)

                        =           0.04 x 10⁵      m/sec

                        =              4 x 10³      m/sec  

                                     (4 kilometers per second)
6 0
3 years ago
A solid disk of mass 2 kg and radius 2 m is given a horizontal push of 20N at a point .3 m above its center. a. What is the mini
Margaret [11]

Answer:

\mu_s=1.0205

Explanation:

Given:

  • mass of solid disk, m=2\ kg
  • radius of disk, r=2\ m
  • force of push applied to disk, F=20\ N
  • distance of application of force from the center, s=0.3\ m

<em>For the condition of no slip the force of  static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

\therefore F

where:

f_s = static frictional force

\Rightarrow 20

\Rightarrow 20

\Rightarrow 20

\mu_s>1.0204

7 0
2 years ago
You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength
Ray Of Light [21]

Answer:

wavelength \lambda = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = \frac{D\lambda}{d}    ....................1

\lambda is Wavelength of light and  D is Distance between screen and slit and d is slit width

so put here value and we get

\lambda = \frac{2.96*0.325*10^{-6}}{2.20}    

\lambda = 437.27 × 10^{-9} m

wavelength \lambda = 437.27 nm

4 0
2 years ago
An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?
sleet_krkn [62]

Answer:

Explanation:

Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:

F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to

F = f which states that the applied Force equals the frictional force, choice a.

3 0
2 years ago
Sound travels at a rate of 340 m/s in all directs through the air. Matt rings a very loud bell at one location, and Steve hears
DENIUS [597]

Answer:

It will take about 1.32 seconds to travel to his location.

Explanation:

Considering the sound travels at 340 m/s, then if a person is at a distance of 450 m m from the bell, we can use the velocity formula to find the answer;

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4 0
3 years ago
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