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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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What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10
raketka [301]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

Y = \frac{FL}{A\Delta L}

20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

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What is transferred to an object when is done A:energy B:force C:effort D:resistance
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Current is constant at all points in a parallel circuit.<br><br> True<br> False
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The baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance, what is th
photoshop1234 [79]

Answer:

9.81 N

Explanation:

Given that a baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance. The maximum height reached will be calculated by using the formula

V^2 = U^2 - 2gH

U = 10 m/s

g = 9.8m/s^2

At maximum height, V = 0

Substitute u and g into the formula

0 = 10^2 - 2 × 9.8 × H

19.6H = 100

H = 100/19.6

H = 5.1 m

The Kinetic energy on the ball will be

K.E = 1/2mv^2

K.E = 1/2 × 1 × 10^2

K.E = 1/2 × 100

K.E = 50 J

But energy = work done

WD = Force × distance (height)

The force that acts on the baseball when it is HALF WAY to the top of the path will be

F × 5.1/2 = 50

F = 100/5.1

F = 19.61 N

The weight acting downward will be

W = mg

W = 1 × 9.8

W = 9.8 N

The net force acting on the ball will be

Net force = F - W

Net force = 19.61 - 9.8

Net force = 9.81 N

Therefore, the net force that acts on the baseball when it is HALF WAY to the top of the path is 9.81 N

4 0
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