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Through how many volts of potential difference must an electron, initially at rest, be accelerated to achieve a wave length of 0

deff fn [24]

Answer:

20.7 volts

Explanation:

m = mass of electron = 9.1 x 10⁻³¹ kg

λ = wavelength of electron = 0.27 x 10⁻⁹ m

v = speed of electron

Using de-broglie's hypothesis

λ m v = h

(0.27 x 10⁻⁹) (9.1 x 10⁻³¹) v = 6.63 x 10⁻³⁴

v = 2.7 x 10⁶ m/s

ΔV = Potential difference through which electron is accelerated

q = charge on electron = 1.6 x 10⁻¹⁹ C

Using conservation of energy

(0.5) m v² = q ΔV

(0.5) (9.1 x 10⁻³¹) (2.7 x 10⁶)² = (1.6 x 10⁻¹⁹) ΔV

ΔV = 20.7 volts

4 0

3 years ago

A football player threw a ball upward at an angle or 24 degrees with the horizontal with A velocity of 18 m/s. What are the vert

svet-max [94.6K]

Answer:

$V_y = 16.44\ m/s$

$V_y = 7.32\ m/s$

Explanation:

Given

$\theta = 24$

$Velocity\ (V) = 18m/s$

Required

Determine the vertical and horizontal components

The vertical (Vy) and horizontal (Vx) components is calculated as thus:

$V_x = Vcos\theta$

$V_y = Vsin\theta$

Calculating Vertical Components:

$V_y = Vsin\theta$

$V_y = 18 * sin24$

$V_y = 18 * 0.40673664307$

$V_y = 7.32125957526$

$V_y = 7.32\ m/s$ --- Approximated

Calculating Horizontal Components:

$V_x = Vcos\theta$

$V_x = 18 * cos24$

$V_x = 18 * 0.91354545764$

$V_x = 16.4438182375$

$V_x = 16.44\ m/s$ --- Approximated

4 0

3 years ago

A 10 kg mass stretches a spring 70 cm in equilibrium. Suppose a 2 kg mass is attached to the spring, initially displaced 25 cm b

NARA [144]

Answer:

Explanation:

1. Find spring constant k. From a free body diagram, you will get the forces on the 10 kg mass, with a displacement d. It will be gravity pulling the mass down and the spring force pulling the mass up. The 10 kg mass is in equilibrium. The resulting equation will be:

$F = m_1a = kd - m_1g = 0 => m_1g = kd => k = \frac{m_1g}{d}$

2. Use the result from 1. to find the equations of motion. In general they are given by:

$x(t) = Acos(\omega t + \phi), v(t) = -\omega Asin(\omega t + \phi)$ , where ω is: $\omega = \sqrt{\frac{k}{m_2}}=\sqrt{\frac{m_1g}{m_2d}}$

To find the amplitude A and the phase angle Ф, use the given initial conditions:

m₂ = 2 kg, x(0) = -0.25 m, v(0) = 2m/s

$-0.25 = Acos(\phi)\\2 = -\omega Asin(\phi)= -\sqrt{\frac{m_1g}{m_2d}}Asin(\phi)\\-0.24 = Asin(\phi)$

Solving for Ф:

$\frac{0.24}{0.25}=tan(\phi) => \phi = 0.76$

Solving for A:

$-0.25 = Acos(\phi) => A = -\frac{0.25}{cos(\phi)}=-0.35$

The equation for x(t) is now:

$x(t) = -0.35 cos(\sqrt{\frac{m_1g}{m_2d}t} +0.76)$

The frequency f is given by: $f = \frac{\omega}{2\pi}$

The period T is given by: $T = \frac{2\pi}{\omega}$

8 0

3 years ago

What is the speed of light m/s

lara31 [8.8K]

299,792,458 m/s in vacuum, somewhat less in any material medium, different in every medium.

3 0

4 years ago

Read 2 more answers

As a student is performing a double slit experiment to determine the wavelength of a light source, she realizes that the nodal l

aleksandr82 [10.1K]

To increase the distance between the nodal lines as it may be difficult to determine the wavelength of the light source because the nodal lines are too close, the student should spread the fringe pattern by having to be able to decrease the separation of the slit.

6 0

3 years ago