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Contact [7]
3 years ago
11

Assume you are riding in a plane that is moving at a speed of 600kmh relative to the ground. If you walk from the back to the fr

ont of the airplane at a speed of 1.0 km/h, then your speed relative to the ground is
A. 1.0 km/h
B. 599 km/h
C. 600 km/h
D. 601 km/h​
Physics
1 answer:
jok3333 [9.3K]3 years ago
5 0
C. 600 km/h is the answer for this question
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Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 542 km above the earth's sur
Evgen [1.6K]
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite 
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g' 
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h): 
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)] 
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec 
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
6 0
3 years ago
Which group of measurements is the most precise? A) 2 g, 3 g, 4 g B) 2 g, 2.5 g, 3 g C) 2.0 g, 3.0 g, 4.0 g, 5.0 g D) 2.0 g, 3.0
siniylev [52]
The answer to your question is A
3 0
3 years ago
a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
MAVERICK [17]

A textbook would hit the ground first


Factors:

-Textbook weighs most

-Pillow is flat and fluffy not very aerodynamic) also is very light

-Paper airplane will glide to the ground do to its wings and will hit the ground last

3 0
3 years ago
Read 2 more answers
:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?​
denis-greek [22]
Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both
3 0
3 years ago
Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a
victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

  1. F=\frac{GMm}{R^{2} }

Where:

  • F => Gravitational force or force of attraction between two masses
  • M => Mass of asteroid 1
  • m => Mass of asteroid 2
  • R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

  • M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

  • R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

  • m=ρV

This is going to be helpful because the volume of a sphere is:

  • \frac{4}{3}\pi r^{3}

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

  • F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }
  • F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }
  • F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2}  }{36}

Now let´s plug in the values we know:

  1. F = 1 lb     mutual gravitational attraction force
  2. G = 6.67(10)^{-11}     gravitational constant
  3. ρ_{iron} =491.5 \frac{lb}{ft^{3} }

  • 1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}

Solve for r and multiply by 2 because 2r = diameter

  • d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }

Result is d = 21.81 Feet

6 0
3 years ago
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