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matrenka [14]
3 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]3 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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The ratio of carbon-14 to nitrogen-14 is an artifact is 1:3. Given that half-life of carbon-14 is 5730years, how old is the arti
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Answer:

9155 years old

Explanation:

We use the following expression for the decay of a substance:

N = N_0\,\,e^{-k*t}

So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:

N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012

so, now we can estimate the age of the artifact by solving for"t" in the equation:

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3 years ago
10.
myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

T'=2\pi \sqrt{\frac{L'}{g}}

T'=2\pi \sqrt{\frac{9L}{g}}

Taking out the square root of 9 (3):

T'=3*2\pi \sqrt{\frac{L}{g}}

Substituting the original T:

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The answer is false.

The force of gravity acting on an object (also known as weight) near the Earth's surface is given by:

F=mg

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m is the mass of the object

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We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

The factor that does not change is the acceleration of gravity, which is constant (9.8 m/s^2) if we are near the Earth's surface, and implies that all the objects in free fall accelerate at the same rate towards the ground, regardless of their size and weight.

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