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3 years ago

5.

Physics

2 answers:

lozanna [386]3 years ago

6 0

Answer:

(1)

Explanation:

iris [78.8K]3 years ago

3 0

Answer:

the answer is a time your welcome

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Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un

Ede4ka [16]

Answer:

What kind of languege is this please?

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2 years ago

A 60 kg hockey player skating at 20 m/s gets slammed with a force of 950 N. How long does it take him to stop completely?

irakobra [83]

Answer:

2.4

Explanation:

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2 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

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

$r_2 = 0.24 \ m$

7 0

3 years ago

Suppose the ends of a 27-m-long steel beam are rigidly clamped at 0°C to prevent expansion. The rail has a cross-sectional area

frozen [14]

Answer:

F = 1.58*10^{11} N

Explanation:

given data:

length of steel beam = 27 m

cross sectional area of rail = 35 cm

$\Delta T = 39$ Degree celcius

change in length of steel beam is given as

$\Delta L = L_O \alpha \Delta T$

$= 20*1.1*10^{-5}*39$

$=8.58*10^{-3}$ m

Young's modulus is

$Y = \frac{FL}{A\Delta L}$

$F = \frac{ YA\Delta L}{L}$

$= \frac{2.0*10^{11}*25*10^{-4}8.58*10^{-3}}{27}$

F = 1.58*10^{11} N

5 0

3 years ago

Takumi works in his yard for 45 minutes each Saturday. He works in the morning, and he wears sunscreen and a hat each time he wo

Arte-miy333 [17]

takumi wants to reduce the chance of getting a sunburn :D

<em>hope this helps!</em>

<em>have a great day :)</em>

6 0

3 years ago

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