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horsena [70]
3 years ago
12

It is the year 802,315 and due to the sun growing, Earth is starting to get too hot for life to survive. Oceans are beginning to

evaporate and soon humanity must move to another planet to survive. Using the machinery below, choose one planet (or moon) in our solar system for humanity to move to and describe specifically how your plan will make that planet (or moon) habitable for the next billion years.
Available Terraforming Machinery

planet tow-ship - this ship can move planets (or moons) smaller than Earth up to 10AU
matter condenser - this device can be sent to another planet (or moon) and turn gas into liquid, and liquids into solids
giant lens - by placing this lens in the just the right location, the heat and light from the sun can be magnified, potentially heating up a planet (or moon)
Assignment Requirements (3-6 sentences):

What planet (or moon) should we colonize?
Why is that planet (or moon) currently not habitable?
How will you use the available machinery to make that planet (or moon) habitable?
What will be different about living on this new planet (or moon) compared to Earth? (for example: Will we need special devices to survive there?)
Physics
1 answer:
Paha777 [63]3 years ago
4 0

Answer:

Explanation:

We will die soon anyway we all die

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half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds
Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %


3 0
3 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
What is the difference between red giant and red dwarf?
Sonbull [250]

the red giant is very tall and the red dwarf is very short.

7 0
3 years ago
Read 2 more answers
The speed of a certain proton is 350 km/s. If the uncertainty in its momentum is 0.100%, what is the necessary uncertainty in it
Evgesh-ka [11]

Answer:

\Delta x = 1.807 \times 10^{-10}m

Explanation:

mass of proton, m = 1.67 x 10^-27 kg

speed of proton, v = 350 km/s = 350,000 m/s

Momentum of proton, p = mass x speed

p =  1.67 x 10^-27 x 350000 = 5.845 x 10^-22 kg m /s

uncertainty in momentum, Δp = 0.1 % of p

Δp = \frac{0.1\times 5.845 \times 10^{-22}}{100}=5.845 \times 10^{-25}

According to the principle

\Delta x\times \Delta p \geq  \frac{h}{2\pi }

where, Δx be the uncertainty in position

\Delta x\times 5.845 \times 10^{-25}=  \frac{6.634 \times 10^{-34}}{2\times 3.14}

\Delta x = 1.807 \times 10^{-10}m

5 0
3 years ago
Calculate the acceleration due to gravity on the moon. the radius of the moon is about 1.74 x 106 m and its mass is 7.35 x 1022
emmainna [20.7K]
Given: Radius of the Moon Rm = 1.74 x 10⁶ m

           Mass of the Moon Mm = 7.35 X 10²² Kg

           Universal Gravitational constant G = 6.67 X 10⁻¹¹ N.m²/Kg²

Required: acceleration due to gravity g = ?



Formula:  g = GMm/Re²

                 g = (6.67 x 10⁻¹¹ N.m²/Kg²)(7.35 x 10²² Kg/(1.74 x 10⁶ m)²

                 g = 4.90 x 10¹² N.m²/Kg/3.0 x 10¹² m²

                 g = 1.63 m/s²




4 0
3 years ago
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