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2 years ago

A machine with a 5-N input force and a 25-N output force has a mechanical advantage of?

1 answer:

8 0

So we want to know the mechanical advantage of a machine that has 5 N input force and 25 N out force. Mechanical advantage Ma is the measure of force amplification of some machine. We calculate it by taking the ratio of the output force Fo over the input force Fi. Ma=Fo/Fi=(25 N)/(5 N)=5. So Mechanical advantage for our machine is Ma=5 and the correct answer is the second one.

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3 years ago

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How do liquids affect friction?

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Although liquids offer resistance to objects moving through them, they also smooth surfaces and reduce friction. Liquids tend to get thinner (less viscous) as they are heated. ... The rubbing produces friction and the result is heat.

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3 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

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3 years ago

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5 0

3 years ago

Read 2 more answers

Two acrobats flying through the air grab and hold onto each other in midair as part of a circus act.One acrobat has a mass of 60

earnstyle [38]

Answer:

1.36m/s

Explanation:

We are given that

Mass of one acrobat, $m_1=60 kg$

Mass of another acrobat, $m_2=50 kg$

$v_2=-3 m/s$

$v_1=5 m/s$

We have to find their velocity immediately after they grab onto each other.

The collision between two acrobat is inelastic

According to law of conservation of momentum

$m_1v_1+m_2v_2=(m_1+m_2)V$

Substitute the values

$60\times 5+50(-3)=(60+50)V$

$300-150=110V$

$V=\frac{300-150}{110}=1.36m/s$

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2 years ago