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Evgen [1.6K]
3 years ago
12

Which hypothenical scenerio would you result in the moon not having dofferent phases? A.The moon takes twice as long as it does

now to orbit the earth. B.Earth takes twice as long as it does now to orbit the sun C Earth has 2 moons. D.The moon always stays in one position relative to the earth and sun. Explain how you got your answer. Who ever answers best gets brainliest!

Physics
1 answer:
maw [93]3 years ago
5 0
<span>The hypothetical scenario that would result in the moon not having different phases would be t</span>he moon always stays in one position relative to the earth and sun. Why? If the moon didn't have an phases it would stay in one position. Phases mean change, and if you observe the moon every night, you'll see how it changes; it repeats, so thus it's a pattern. 

So, your correct answer should be: <span>D.The moon always stays in one position relative to the earth and sun.</span>
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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}}))

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Class 1 lever

Explanation:

In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.

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Load related problems brainly.com/question/9202964

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