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3 years ago

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What will happen to the wavelength of light uf the frequency is doubled?What will happen to the wavelength of light uf the frequ

ency is doubled?

1 answer:

otez555 [7]3 years ago

6 0

Answer:

if the frequency is double, the wavelength is only half as long

Explanation:

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Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o

Harman [31]

Answer:

A) 1.55

B) 1.55

C) 12.92

D) 34.08

E) 57.82

Explanation:

The free body diagram attached, R is the radius of the wheel

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.

At the centre of the whee, torque due to B is given by

${\tau _2} = - {T_{\rm{B}}}R$

Similarly, torque due to A is given by

${\tau _1} = {T_{\rm{A}}}R$

The sum of torque at the pivot is given by

$\tau = {\tau _1} + {\tau _2}$

Replacing ${\tau _1}$ and ${\tau _2}$ by ${T_{\rm{A}}}R$ and $- {T_{\rm{B}}}R$ respectively yields

$\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}$

Substituting $I\alpha$ for $\tau$ in the equation $\tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

The angular acceleration of the wheel is given by $\alpha = \frac{a}{R}$

where a is the linear acceleration

Substituting $\frac{a}{R}$ for $\alpha$ into equation

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ we obtain

$\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

Net force on block A is

${F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}$

Net force on block B is

${F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g$

Where g is acceleration due to gravity

Substituting ${m_{\rm{B}}}a$ and ${m_{\rm{A}}}a$ for ${F_{\rm{B}}}$ and ${F_{\rm{A}}}$ respectively into equation $\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ and making a the subject we obtain

$\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}$

Since ${m_{\rm{B}}}$ = 3kg and ${m_{\rm{B}}}$ = 7kg

g=9.81 and R=0.12m, I=0.22 ${\rm{ kg}} \cdot {{\rm{m}}^2}$

Substituting these we obtain

$a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}$

$\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the linear acceleration of block A is 1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(B)

For block B

${a_{\rm{B}}} = {a_{\rm{A}}}$

Therefore, the acceleration of both blocks A and B are same

1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(C)

The angular acceleration is $\alpha = \frac{a}{R}$

$\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

(D)

Tension on left side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}$

(E)

Tension on right side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}$

6 0

3 years ago

Imagine two fixed charges on the x axis. Charge one is +q and is located to the left of charge two which is equal to -4q. Where

givi [52]

Answer: B)To the left of the charges.

Explanation: between the charges the electric field will not cancel but will be added since electric field lines from both charges point in the same direction. To the right of the charge the -4q will take over as it’s strength overcomes the strength of the +q charge. At this point the magnitude of +q will never reach a magnitude strong enough to cancel the -4q. To the left, it is further away from -4q and is closer to +q and electric field lines point in different direction

7 0

3 years ago

A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr

mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

$f' = (\frac{v}{v\pm v_s})f$

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

$v_s$ is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

$v_s = -40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz$

When the train has passed the platform, we have

$v_s = +40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz$

Therefore the overall shift in frequency is

$\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz$

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

$v=\lambda f$

where

v is the speed of the wave

$\lambda$ is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

$\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m$

5 0

3 years ago

What occurs when the pitch increases and the sound source is in motion? explain your answer.

Hoochie [10]

If you're listening to a sound that has a steady pitch, and suddenly the
pitch goes up, then you know that two things could have happened:

EITHER ...

-- The person or other source making the sound could have
raised the pitch of the sound being produced.

OR ...

-- The person or other source making the sound could have
started moving toward you.

OR ...

-- both.

Even if the pitch of the sound leaving the source doesn't change,
you would still hear it increase if the source starts moving toward
you. That's the so-called "Doppler effect".

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3 years ago

For pea plants, T represents a dominant allele for tall pea plants and t is the recessive allele for short pea plants. If two pl

JulsSmile [24]

Answer:

75%(TT,Tt,tt) Hope that helps

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2 years ago