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tensa zangetsu [6.8K]
4 years ago
6

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.
1) What would be the acceleration, in m/s, of such a rock fragment?
2) What would be the acceleration, in g's, of such a rock fragment?
3) How long would this acceleration last?
4) In tests, scientists have found that over 40% of Bacillius subtilis bacteria survived after an acceleration of 450000g. In light of your answer to part A, can we rule out the hypothesis that life might have been blasted from Mars to Earth?
Physics
1 answer:
aksik [14]4 years ago
4 0

Answer and Explanation:

Given that

v_f = 5 km/s = 5,000 m/s

d = 4 m

v_i = 0 m/s

The computation is shown below:

1. The acceleration in m/s is

Here we use the motio third equation which is

v_f^2 = v_i^2 + 2ad

5000^2 = 0^2 + 2 (a) (4)

So

a = 3.125 \times 10^{6} m/s^2

2. Now acceleration in g is

= \frac{3.125 \times 10^{6} m/s^2}{9.81}

= 3.18 \times 10^{5}g

3. The long of acceleration last is

t = \frac{v-u}{a}

= \frac{5000 - 0}{3.125 \times 10^{6}}

= 1.6 \times 10^{-3}s

4.As we can see that

3.18 \times 10^{5} is smaller than the 4.5 \times 10^{5}g

So, it should not be ruled out

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Solve using correct significant figures and indicating maximum absolute uncertainty.
Vera_Pavlovna [14]

We use the criterion of significant figures to find the result with reliable figures

          X = 9.2 10-5

now with the propagation of errors we obtain the result with its uncertainty

         X ± ΔX = (9.2 ± 0.5) 10⁻⁵

given Parameter

     * expression values ​​with their absolute errors

to find

     * the result with the correct significant figures

     * the absolute error of the expression

Significant figures are defined with the number of decimals that give information, the number of figures in a quantity gives information about the uncertainty of this quantity.

There are two criteria for applying significant figures:

     * Add and subtract the result of going with the number of decimal places of the figure that has the least

    * Product and division as a result of going with the least number of significant figures than the value that has the least.

Remember that the zero to the left do not form a pair of the significant figures

Let's apply this belief to the case presented, let's write the precaution

 

              x = \frac{a-b}{c}

where in this case they are worth

         a = 0.0336 ± 0.0002

         b = 0.010 ± 0.001

         c = 255.4 ± 0.4

We see that the significant figures of each parameterize (a, b, c) and their absolute errors are correct.

Let's apply the criteria to the operation

          a-b = 0.0336 - 0.010

          a- b = 0.0236

we apply the criterion of significant figures for the subtraction, the result must be with 3 decimal places

        a - b = 0.024

let's do the other operation

         X = \frac{a-b}{c}

         X = 0.024 / 255.4

         X = 9.24 10⁻⁵

We apply the criterion of significant figures for the division, in this case the result is left with two significant figures

         X = 9.2 10⁻⁵

The uncertainty or error of the measurements is of most importance as it determines how many significative figures are reliable at a given magnitude.

If the magnitudes are measured with some type of instrument, the absolute error is given by the appreciation of the instrument, if the magnitude is calculated using some equation, the errors must be propagated using the variations of each parameter in the worst case.

           

the uncertainty of the calculated quantity (X) is

        \Delta X = | \frac{dX}{da}| \Delta a + | \frac{dX}{db} | \Delta b + | \frac{dX}{dc}| \Delta c

let's perform the derivatives

        \frac{dX}{da} = \frac{1}{c}

        \frac{dX}{db} = - \frac{1}{c}

        \frac{dX}{dc} = - \frac{a-b}{c^2}

we substitute

remember that the bulk value guarantees that we tune the worst case. So all the mistakes add up

          ΔX = \frac{1}{c}  Δa + \frac{1}{c} Δb + \frac{a-b}{c^2}  Δc

          ΔX = \frac{1}{c} (Δa + Δb) + \frac{a-b}{c^2} Δc

we substitute

         ΔX = \frac{1}{255.4}  (0.0002 + 0.001) + \frac{0.0336-0.010}{255.4^2}  0.4

         ΔX = 4.698 10⁻⁶ + 1.45 10⁻⁷

         DX = 4.8 10-6

Absolute errors must be given with a single significant figure

         ΔX = 5 10⁻⁶

The result of the requested quantity using the criterion of significant figures and propagation of errors is

          X ± ΔX = (9.2 ± 0.5) 10⁻⁵

learn more about   significative figure here:

brainly.com/question/18955573

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According to the law of conservation of matter, the number of ________ is not changed by a chemical reaction.
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The answer should be atoms.
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1) Find the voltage on a circuit with a resistance of 12.5 Ω if it has a current of 2.35 A.
KengaRu [80]

1. The voltage on the circuit with a resistance of 12.5 Ω and current of 2.35 A is 29.4 V.

2.The resistance in the circuit is found to be 1.45 Ω.

3. The equivalent resistance of the resistors connected in series is 24 Ω.

4. The equivalent resistance of the resistors connected in parallel is 2.18Ω.

5. The power generated is 552.5 W.

6. The frequency of the green light is 0.56×10¹⁵ Hz.

Explanation:

1) This problem can be solved using Ohm's law. Here the resistance (R) of the circuit is given to be 12.5 Ω and the current (I) is stated to be 2.35 A. So the ohm's law states that in a closed circuit, the voltage will be directly proportional to the current flowing in the circuit and the resistance will act as the proportionality constant.

V = I * R = 2.35*12.5 = 29.4 V

So, the voltage on the circuit with a resistance of 12.5 Ω and current of 2.35 A is 29.4 V.

2) Using the same Ohms' law, now we have to determine the resistance. So in this case, the voltage is given as 9 V and the current is said to be 6.2 A, then resistance can be determined as the ratio of voltage to current.

R = \frac{V}{I} =\frac{9}{6.2} =1.45 Ohms

So, the resistance in the circuit is found to be 1.45 Ω.

3) Here, the resistances of three resistors are given as 4 Ω, 8 Ω and 12 Ω. And it is stated that the resistances are connected or wired in series. Then the equivalent resistance will be obtained by the sum of resistances of three resistors, as the current flow will be constant in all the three resistors.

R_{s} = R_{1} + R_{2} + R_{3}  \\  \\R_{s} = 4+8+12 = 24 ohms

Thus, the equivalent resistance of the resistors connected in series is 24 Ω.

4) Now, if the resistors are connected in parallel, then the equivalent resistance will be ratio of product of resistances to the sum of the resistances.

\frac{1}{R_{p} }= \frac{1}{R_{1} } + \frac{1}{R_{2} } +\frac{1}{R_{3} }\\\\\frac{1}{R_{p} }=\frac{1}{4}+ \frac{1}{8} +\frac{1}{12} = \frac{6+3+2}{24}  =\frac{11}{24} \\\\R_{p} = \frac{24}{11 } =2.18 Ohm

Thus, the equivalent resistance of the resistors connected in parallel is 2.18Ω.

5) Power generated by the person can be obtained by the ratio of work done by the person to the time in which the work is done. So the work done can be obtained by the product of force with displacement.

As here the weight lifted by the person will act as dominant force on the person. So the force is considered as F = 956 N and the displacement is d = 2.41 m, then

Work done = Force * displacement = 956*2.41 =2303.96 J

So, the work done is obtained as 2303.96 J and the time is given as 4.17 s, then

Power = \frac{Work done}{Time} =\frac{2303.96}{4.17} =552.5 W

So, the power generated is 552.5 W.

6) In this, the wavelength of green light is given as 5.34 × 10⁻⁷ m. It is known that the wavelength is inversely proportional to the frequency.

Wavelength = \frac{Speed of light}{Frequency}

As, speed of light is known as 3×10⁸ m/s, the frequency will be determined as the ratio of speed of light to wavelength.

Frequency = \frac{Speed of light}{Wavelength} =\frac{3*10^{8} }{5.34*10^{-7} } \\\\Frequency =0.56*10^{15} Hz

Thus, the frequency of the green light is 0.56×10¹⁵ Hz.

8 0
4 years ago
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