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3 years ago

8

Choose the word that BEST completes the following sentence. Saturn’s rings have been assigned a letter, based on their location.

The rings A and B are separated by: Titan The Cassini Division Stars An open space fairly equivalent to the Grand Canyon

1 answer:

puteri [66]3 years ago

5 0

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

https://caps.gsfc.nasa.gov/simpson/kingswood/rings/

Hope this helps! :)

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A beam of electrons with of wavelength of 7.5 x 10-6 m is incident on a pair of narrow rectangular slits separated by 0.75 mm. T

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4 years ago

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

where n=1,2,3 or ...n

L=Length of string

T=Tension

$\mu =$ Mass per unit length

When string is clamped at mid-point

Effecting length becomes $L'=0.5 L$

Thus new Frequency becomes

$f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}$

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz

4 0

3 years ago

If a ball is tossed straight up into the air, at what position is its potential energy the greatest? Question 2 options: a) when

yarga [219]

Q1: At the highest point

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4 years ago

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

4 years ago

A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is

Sonja [21]

Answer:

0.8895m

Explanation:

Cable diameter = 0.0125m

Mass of elevator = 6450kg

Young Modulus(E) = 2.11*10¹¹N/m

∇l (change in length) =

L = 362m

A = Πr², but r = d / 2 = 0.0125 / 2 = 0.00625m

A = 3.142 * (0.00625)² = 1.227*10^-4m²

Young Modulus (E) = Tensile stress / Tensile strain

E = (F / A) / ∇l / L

F = mg = 6450 * 9.8 = 63210N

2.11*10¹¹ = (63210 / 1.22*10^-4) / (∇l / 362)

2.11*10¹¹ = 5.18*10⁸ / (∇l / 362)

2.11*10¹¹ = (5.18*10⁸ * 362) / ∇l

2.11*10¹¹ = 1.875*10¹¹ / ∇l

∇l = 1.875*10¹¹ / 2.11*10¹¹

∇l = 0.8895m

The change in length is 0.8895m

8 0

3 years ago