The main equation we are using to measure the e/m ratio is:

A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen

Brums [2.3K]

We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
$\frac{h_i}{h_0}=\frac{d_i}{d_0}$
Two triangles that I marked are similar and from this we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f}$
The image and the object must have the same height so we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f$
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
$S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm$

8 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

A photon is scattered from an initially stationary electron within a metal. How does the frequency of the photon change upon sca

sammy [17]

Answer:

The frequency of the photon decreases upon scattering

Explanation:

Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.

Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.

Therefore, since the wavelength increases, we have from

λf = λ'f' = c

f = c/λ

Where:

f and f' = The frequency of the motion of the photon before and after the scattering

c = Speed of light (constant)

We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;

f = c/λ

As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.

5 0

2 years ago

The actions of an employee are not attributable to the employer if the employer has not directly or indirectly encouraged the em

zepelin [54]

Answer: the answer is d

Explanation: there are not more than 10 violations within a twelve month period hope this helps

4 0

2 years ago

An x-ray tube is an evacuated glass tube that produces electrons at one end and then accelerates them to very high speeds by the

laila [671]

Answer:

a) V = 1.866 10² V , b) V = 3.424 10⁵ V , c) v = 8.1 10⁶ m / s

Explanation:

a) the potential difference is requested to accelerate the electrons up to 2.7% of the speed of light

v = 0.027 c

v = 0.027 3 10⁸

v = 8.1 10⁶ m / s

for this part we can use the conservation of mechanical energy

starting point. When electrons are at rest

Em₀ = U = q V

final point. Electrons with maximum speed

Em_f = K = ½ m v2

Em₀ = Em_{f}

e V = ½ m v²

V = ½ m v² / e

let's calculate

V = ½ 9.1 10⁻³¹ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 1.866 10² V

V = 1866 V

b) if this acceleration protons is the mass of the proton is m_{p} = 1.67 10-27

V = ½ 1.67 10⁻²⁷ (8.1 10⁶)² / 1.6 10⁻¹⁹

V = 3.424 10⁵ V

V = 342402 V

c) this potential difference should give the protons the same speed as the electrons

v = 8.1 10⁶ m / s

5 0

2 years ago