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vladimir2022 [97]
2 years ago
11

Suppose we have a space ship orbiting the earth. What must happen in order for the spaceship to leave orbit and fall back toward

earth?
The spaceship will fall back toward earth on its own but you have to just wait for it to fall.

The spaceship's forward motion must be slowed down so the earth's gravitational pull on it will be stronger than the ship's forward motion.

The spaceship's forward motion must be sped up so the earth's gravitational pull on it will be weaker than the ship's forward motion.

It is not possible for a spaceship orbiting the earth to leave orbit and fall back to earth once it is in orbit.
Physics
2 answers:
professor190 [17]2 years ago
8 0
It’s option 2
Hope it will help you:)
sdas [7]2 years ago
3 0

Answer:

#2) The spaceship's forward motion must be slowed down so the earth's gravitational pull on it will be stronger than the ship's forward motion.

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four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?
Kay [80]
There was no net force on the stuffed toy, because the kids might have the same strength,  The same force is on both sides of it.  T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move.  </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving. 
3 0
3 years ago
Read 2 more answers
Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27
densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

\rho=n\times\dfrac{m}{V_{c}\times N_{A}}

n=\dfrac{\rho\timesV_{c}\times N}{m}....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100

\rho=5.98

We calculate the mass of the alloy

m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100

m=111.15

Put the value into the equation (I)

n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}

n=1.99\approx 2\ atoms/cell

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0
4 years ago
What is the magnitude of the sum of the two vectors A = 36 units at 53 degrees, and B =47 units at 157 degrees.
Thepotemich [5.8K]

Answer:

51.82

Explanation:

First of all, let's convert both vectors to cartesian coordinates:

Va = 36 < 53° = (36*cos(53), 36*sin(53))

Va = (21.67, 28.75)

Vb = 47 < 157° = (47*cos(157), 47*sin(157))

Vb = (-43.26, 18.36)

The sum of both vectors will be:

Va+Vb = (-21.59, 47.11)   Now we will calculate the module of this vector:

|Va+Vb| = \sqrt{(-21.59)^2+(47.11)^2}=51.82

4 0
4 years ago
The period and freqency of a wave are inversely related true or false​
Soloha48 [4]

Answer: false

Explanation: the longer the period, the less thef= frequency

5 0
3 years ago
Read 2 more answers
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
matrenka [14]

Answer:

a) 6.1 m

b) 4.6 s

c) 1.326 m/s

d) 0.325 m

Explanation:

a) The wave length is the distance between 2 crests λ = 6.1m

b) The period of the wave is the time it takes from the lowest point to the next lowest point, which is twice the time it takes from the lowest point to the highest point = 2*2.3 = 4.6 s

c) The speed of the wave is the distance per unit of time, or wave length over period = 6.1 / 4.6 = 1.326 m/s

d)The amplitude A is half the distance from the highest point to the lowest point = 0.65 / 2 = 0.325 m

6 0
3 years ago
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