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Mariulka [41]
3 years ago
11

Select all of the answers that apply.

Physics
2 answers:
Kruka [31]3 years ago
6 0
The answer is letter A. meteorite bombardment.
During the Earth's earliest beginning, it went through a period of catastrophic and intense formation. By 3-8- 4.1 billion years ago, Earth's atmosphere was never the same as today. This is because of its formation during the pre-Cambrian period whereby t<span>he Earth formed under so much heat and pressure that it formed as a molten planet.
</span>

Earth was bombarded continuously by the remnants of the dust and debris — like asteroids, meteors and comets — until it formed into a solid sphere, pulled into orbit around the sun and began to cool down during the Hellish period (4.5 to 3.8 billion years ago).

<span> </span>.To know more of this topic, see attached file.

 

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Helga [31]3 years ago
4 0

Answer: meteorite bombardment

Explanation: This is because of its formation during the pre-Cambrian period whereby the Earth formed under so much heat and pressure that it formed as a molten planet.

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One of two 25-year-old identical twins begins a trip on a spaceship traveling at 0.8 c while her twin remains on Earth. The twin
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2 years ago
Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative
SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

Explanation:

Given there are three blocks of masses M_{a}, M_{b} and M_{c} (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F =  (M_{a} + M_{b} + M_{c} ) *a    ................(equation 1)

Also it is given that M_{a} does not move with respect to M_{c}, which gives tension T  is exerted on pulley  by M_{a} only, Hence tension T is

T = M_{a} *a    ..........(equation 2)

There is also also tension exerted by M_{b}. There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = M_{b} \sqrt{a^{2} +g^{2} }   ................(equation 3)

From equation 2 and 3, we get

M_{a} *a  = M_{b} \sqrt{a^{2} +g^{2} }  

Squaring both sides we get

M_{a} ^{2} *a^{2} = M_{b} ^{2} * (a^{2}+g^{2})

M_{a} ^{2} *a^{2} = (M_{b} ^{2} * a^{2})+ (M_{b} ^{2} *g^{2})

(M_{a} ^{2}  -  M_{b} ^{2}) * a^{2} = M_{b} ^{2} *g^{2}

a^{2} = M_{b} ^{2} *g^{2}/(M_{a} ^{2}  -  M_{b} ^{2})

Taking square root on both sides, we get acceleration a

a = M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}})

Hence substituting the value of a in equation 1, we get

F =  (M_{a} + M_{b} + M_{c} ) *(M_{b}*g/(\sqrt{M_{a} ^{2}-M_{b} ^{2}}))

3 0
3 years ago
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