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Paul [167]
3 years ago
12

An astronaut on the moon pushes with her left hand on a small rock and with her right hand on a large rock. Each hand pushes wit

h a force of 15 N. The small rock has mass 2 kg and the large rock has mass 8 kg. What forces do the rocks exert on the astronaut?
The small rock exerts a pushing force of 20 N; the large rock exerts a pushing force of 78 N.

The small rock exerts a pushing force of 3 N; the large rock exerts a pushing force of 12 N.

Each rock exerts a pushing force of 7.5 N.

Each rock exerts a pushing force of 15 N.
Physics
2 answers:
trapecia [35]3 years ago
7 0

Both would go 15N. Hoped I helped!

Marina CMI [18]3 years ago
4 0

Answer:

<h2>Each rock exerts a pushing force of 15 N.</h2>

Explanation:

The clue here is that the astronaut pushes with a force of 15N.

To understand this better we need to recur to Newton's third law which states

For every action (force), there's an equal and opposite reaction.

So, we know that the astronaut pushed both rocks with 15N force. By Netwon's third law, both rocks exert a force of 15N back, because such reaction must be equal.

Therefore, the right answer is the last choice: <em>Each rock exerts a pushing force of 15 N.</em>

<em />

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Answer:

a. If an object's speed is constant, then its acceleration must be zero.

FALSE

As we know that acceleration is defined as the rate of change in velocity

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b. If an object's acceleration is zero, then its speed must be constant.

TRUE

As we know that acceleration is defined as the rate of change in velocity

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Since we know that if acceleration is 0 then velocity must be constant and hence speed is also constant

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d. If an object's acceleration is zero, its velocity must be constant.

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7 0
2 years ago
You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30
12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

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using formula

\beta = 10 log(\dfrac{I_1}{I_0})

   I₀ = 10⁻¹² W/m²

now,

30 = 10 log(\dfrac{I_1}{10^{-12}})

\dfrac{I_1}{10^{-12}}= 10^3

I_1= 10^{-8}\ W/m^2

to hear the whisper sound = 80 dB

80 = 10 log(\dfrac{I_2}{10^{-12}})

\dfrac{I_2}{10^{-12}}= 10^8

I_2= 10^{-4}\ W/m^2

we know intensity of sound is inversely proportional to square of distances

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}

\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}

10^{-4}=\dfrac{r_2^2}{20^2}

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