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bekas [8.4K]
3 years ago
6

A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electr

ons and the protons have the same speed. Which one of the following correctly describes the beam that experiences the greatest amount of diffraction?
(a) The electrons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
(c) The protons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(d) The protons, because they have the larger momentum and, hence, the smaller de Broglie wavelength
(e) Both beams experience the same amount of diffraction, because the electrons and protons have the same de Broglie wavelength.
Physics
1 answer:
ryzh [129]3 years ago
7 0

Answer:

(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength

Explanation:

de Broglie wavelength λ = h / m v

Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .

for electron , momentum is less so  . Therefore de Broglie wavelength λ will be more for electrons .

Amount of diffraction that is angle of diffraction is proportional to λ

Therefore electrons having greater de Broglie wavelength will show greater diffraction.

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What is the restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b? A. –7 ab B. `-7 a/b ` C
algol13

Answer:

C. -12 ab

Explanation:

The restoring force on a spring is given by Hooke's law:

F=-kx

where

k is the spring constant

x is the stretched (or compressed) displacement of the spring

In this problem we have:

k = 4a

x = 3b

Substituting into the equation, we find:

F=-(4a)(3b) = -12 ab

And the negative sign means that the direction of the force (negative) is opposite to the direction of the displacement (positive).

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3 years ago
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west
Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

v_B=+5\ m/sis the velocity of Bill with respect to Amy (which is stationary)

v_c=15\ m/s is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

v_B=+5\ m/s

Therefore, Carlos velocity in Bill's reference frame will be

v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s

So, the magnitude is 20 m/s and the direction is westward (negative sign).

7 0
3 years ago
A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= 10+1.50

= 11.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{11.5}

      =26.08 \ s

Throughout the opposite direction, when the boat seems to be travelling then,

= 10-1.50

= 8.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{8.5}

      =35.29 \ s

hence,

The time taken by the boat will be:

= 26.08+35.29

= 61.37 \ s

8 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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3 years ago
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