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matrenka [14]
2 years ago
8

It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit

Mars and blasted pieces of rock (perhaps containing primitive life) free of the surface. Astronomers know that many Martian rocks have come to Earth this way. One objection to this idea is that microbes would have to undergo an enormous, lethal acceleration during the impact. Let us investigate how large such an acceleration might be. To escape Mars, rock fragments would have to reach its escape velocity of 5.0 km/s, and this would most likely happen over a distance of about 4.0m during the impact.
What would be the acceleration, in m/s, of such a rock fragment?
Physics
1 answer:
damaskus [11]2 years ago
5 0

Answer:

a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

Explanation:

According to Newton's Third Equation of motion

V_f^2-V_i^2=2as

Where:

V_f is the final velocity

V_i is the initial velocity

a is the acceleration

s is the distance

In our case:

V_f=V_{escape},  V_i=0,s=4 m

So Equation will become:

V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

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Acceleration(a) is the change in velocity(Δv) over the change in time(Δt). so just divide your velocity and time.
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2 years ago
The greater the of an object the more force is needed to cause acceleration
d1i1m1o1n [39]

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8 0
3 years ago
You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro
guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0
3 years ago
Read 2 more answers
Two poles are connected by a wire that is also connected to the ground. The first pole is 20ft tall and the second pole is 10ft
Natasha2012 [34]

Answer: 31.6ft

Explanation:

Check the attachment for the diagram.

According to the right angle triangle AEC, we will use Pythagoras theorem to get |AC|. Note that |AE| = |AB| - |CD|

that is 20ft - 10ft = 10ft

According to the theorem, the square of the sum of the adjacent side and the opposite side is equal to the square of the hypotenuse.

|AE|^2 + |EC|^2 = |AC|^2

10^2 + 30^2 = |AC|^2

100 + 900 = |AC|^2

|AC| = √1000

|AC| = 31.6ft

Therefore, the wire should be anchored 31.6ft to the ground to minimize the amount of wire needed.

6 0
3 years ago
Why might it be necessary to ignore some of the data points just before and just after the collision?
krek1111 [17]
We have no idea. We need to examine the experimental set-up. You've given us no information, except that there may have been some sort of collision.
7 0
3 years ago
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