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jasenka [17]
3 years ago
7

If a cat falls off a ledge, the Earth pulls the cat to the ground with the force of gravity. According to Newton's Third Law the

re must be a reaction force. What is that reaction force?
A. The cat pulls on the Earth
B. The cat’s acceleration is decreased by air resistance
C. The Earth hits the cat
D. The cat falls to the ground
E. The cat hits the Earth
Physics
2 answers:
elena55 [62]3 years ago
8 0

Answer:

<em><u>The</u></em><em><u> </u></em><em>answer</em><em> </em><em>Is</em><em> </em><em>B.</em><em> </em>

Explanation:

<em><u>The</u></em><em><u> </u></em><em><u>cats</u></em><em><u> </u></em><em><u>acceleration</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>decreased</u></em>

<em><u>by</u></em><em><u> </u></em><em><u>air</u></em><em><u> </u></em><em><u>resistance</u></em><em><u>.</u></em><em><u> </u></em>

<em><u>hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u> </u></em><em><u>you</u></em><em><u>!</u></em><em><u> </u></em>

<em><u>follow</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>more</u></em>

<em><u>don't</u></em><em><u> </u></em><em><u>worry</u></em><em><u> </u></em><em><u>I'll</u></em><em><u> </u></em><em><u>try</u></em><em><u> </u></em><em><u>my</u></em><em><u> </u></em><em><u>best</u></em>

Andru [333]3 years ago
5 0
THE
ANSWER
TO
UR
QUESTION
IS
B
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Answer:

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Answer:

See Explanation

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From the question, we understand that:

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t \to T --- time

Remove the other terms of the equation, we have:

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s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

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s=v_0t

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v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

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v_0 = LT^{-1}

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Solving (c): Units and dimension of a_0

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s=\frac{a_0t^2}{2}

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\frac{a_0t^2}{2} = s_0

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s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

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Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

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Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

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Make S_0 the subject

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Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

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S_0 = LT^{-4 --- dimension

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Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

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Rewrite as:

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c = \frac{L}{T^5}

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Hence:

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