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3 years ago

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If the strings have the same thickness butdifferent lengths, which of the following parameters, if any, willbe different in the

two strings?Check all that apply.wave frequencywave speedwavelengthnone of the above

1 answer:

gizmo_the_mogwai [7]3 years ago

7 0

Answer:

1.Length is one of the four factors on which the wave frequency depends. So if the length of the string changes then there will be a change in the vibration of string. So in this case if the lengths are different then the wave frequency of both will be different.

2. Wave speed will be the same as it depends on tension and linear density of it.

3. Wavelength itself is find out by the length of string so it depends on length and it will vary with the lengths of strings.

Explanation:

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe

Ede4ka [16]

The final velocity ( $v_1_f$ ) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.

The given parameters;

Mass of the first astronaut, = m₁
Mass of the second astronaut, = m₂
Initial velocity of the first astronaut, = v₁
Initial velocity of the second astronaut, = v₂ > v₁
Mass of the ball, = m
Speed of the ball, = u
Final velocity of the first astronaut, = $v_f_1$
Final velocity of the second astronaut, = $v_f_2$

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

$m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f$

if v₂ > v₁, then $v_1_f > v_2_f$ , to conserve the linear momentum.

Thus, the final velocity ( $v_1_f$ ) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut ( $v_2_f$ ) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

5 0

2 years ago

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain:

$\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}$

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0

3 years ago

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b

lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting frictional force is provided by Centripetal force

$f=\mu N=\mu m\omega ^2\cdot r$

weight of object=mg

Equating two

f=mg

$\mu m\omega ^2\cdot r=mg$

$\omega ^2=\frac{g}{\mu r}$

$\omega =\sqrt{\frac{g}{\mu r}}$

$\omega =2.55 rad/sec$

8 0

3 years ago

Is a process that modifies light waves so they vibrate in a single plane.

Naddik [55]

You're fishing for "polarization".

4 0

3 years ago

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