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2 years ago

A compass in a magnetic field will line up __________.

Physics

1 answer:

Klio2033 [76]2 years ago

4 0

Answer:

Option c

Explanation:

Magnetic field lines form loops starting from north pole to south pole outside the magnet and from south pole to north pole inside the magnet.

Thus the field is such that it is directed outwards from the North pole and directed inwards to the South pole of the magnet.

A compass in a magnetic field will will comply with the magnet's North pole directing towards the magnetic field.

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Which letter (A, B, or C) shows where you should apply the most effort to lift the stone?

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2 years ago

What speed must an electron have if its momentum is to be the same as that of an x-ray photon with a wavelength of 0.30?

lorasvet [3.4K]

The momentum of the x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and h is Planck's constant,(h=6.62607004 × 10-34 m2 kg / s).The momentum is: 2.2 x 10^(-25).

The momentum can be calculated also as: p=mv, where m is the mass of the electron and v is the speed.

So v=p/m,p is known,and also the mass of the electron (m=9.10938356 × 10-31 kilograms).

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3 years ago

An object that is falling has the following type(s) of energy. Ignore air resistance.

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2 years ago

Read 2 more answers

A sound has a higher pitch than a bell. Which quantity must also be higher in this sound than in the bell?

Talja [164]

Answer:

frequency

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