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Lilit [14]
3 years ago
13

When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not n

oticeable?​
Physics
2 answers:
Anarel [89]3 years ago
6 0

Answer:

True, because unlike the apple we don't have a large as$ refrence point (the earth is too big to notice being pushed)

Explanation:

Katena32 [7]3 years ago
4 0

Yes, this is true.

-- While the apple is falling, the same gravitational force acts on both the apple and the Earth.

-- The mass of the apple is somewhere in the neighborhood of 1/4 kg.

-- The mass of the Earth is about 5.972 x 10²⁴ kg.

-- Since the Earth has roughly 2.389 x 10²⁶ times as much mass as the apple has, the apple has roughly 2.389 x 10²⁶ greater acceleration than the Earth has, and moves roughly 2.389 x 10²⁶ times as far down as the Earth moves up, before they smack together.

-- That's why you don't notice the Earth's motion.

-- Also, you're standing on the Earth, moving up with it, toward the apple.  Maybe it would be different if you were sitting on the apple, riding it down to the ground, and you were able to notice the motion of the ground coming up to meet you at a speed that's 0.00000000000000000000000000419 of YOUR speed.  

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Answer:

1.55 N

Explanation:

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m = 157.8 g

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W = (0.1578 kg) (9.8 m/s²)

W = 1.55 N

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Which of the below statements are true?
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Answer:

B

Explanation:

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if you start at point A, then go to point B, and back to point A, the displacement is zero because you started and ended at the same point.

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A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

a) Rate=k[A]^1

k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

8 0
3 years ago
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