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1 year ago

6

When a beam of charged particles moves through a magnetic field, what is the evidence that particles in the beam have momentum g

reater than the value mv?

1 answer:

Vsevolod [243]1 year ago

4 0

The charged particles are often deflated in a magnetic field.

<h3>What is a magnetic field?</h3>

The term charge refers to a positive or negative entity. The can be created when a charge is made to pass through a conductor in a magnetic field.

A magnetic field is created when we have a north pole and a south pole. The charged particles could be made to pass through the electric field and when that happens, we can see a pattern a shown in the image attached.

Thus, we can see that the charged particles are often deflated in a magnetic field.

Learn more about magnetic field:brainly.com/question/23096032

#SPJ1

You might be interested in

kipiarov [429]

<h2>Answer: Diamond</h2>

Explanation:

This described situation is known as Refraction, a phenomenon in which the light bends or changes it direction when passing through a medium with a index of refraction different from the other medium.

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium index of refraction (air)

$n_{2}$ is the second medium index of refraction (the value we want to know)

$\theta_{1}=27\°$ is the angle of the incident ray

$\theta_{2}=11\°$ is the angle of the refracted ray

Now, let's find $n_{2}$ from (1):

$n_{2}=n_{1}\frac{sin(\theta_{1}}{sin(\theta_{2}}$ (2)

Substituting the known values:

$n_{2}=(1)\frac{sin(27\°)}{sin(11\°)}}$

Finally:

$n_{2}=2.379\approx 2.4$

If we compare this result with the given table, the index of refraction value that is close to this number is diamond's index of refraction.

Therefore, the correct option is A: the material is diamond.

3 0

3 years ago

Please can someone solve this physics question with a good explenation.

zimovet [89]

Answer:

The coefficient of dynamic friction is 0.025.

Explanation:

Given:

Initial speed after the push is 'v' as seen in the graph.

Final speed of the stone is 0 m/s as it comes to rest.

Total distance traveled is, $D=29.8\ m$

Total time taken is, $t_{total}=17.5\ s$

Time interval for deceleration is 3.5 to 17.5 s which is for 14 s.

Now, average speed of the stone is given as:

$v_{avg}=\frac{D}{t_{total}}=\frac{29.8}{17.5}=1.703\ m/s$

Now, we know that, average speed can also be expressed as:

$v_{avg}=\frac{v_i+v_f}{2}\\1.703=\frac{v+0}{2}\\v=2\times 1.703=3.41\ m/s$

Now, from the graph, the vertical height of the triangles is, $v=3.41\ m/s$

The deceleration is given as the slope of the line from time 3.5 s to 17.5 s.

Therefore, deceleration is:

$a=\frac{\textrm{Vertical height}}{\textrm{Time interval}}\\a=\frac{v-0}{17.5-3.5}\\a=\frac{3.41}{14}=0.244\ m/s^2$

Frictional force is the net force acting on the stone. Frictional force is given as:

$f=\mu_dN\\Where, \mu_d\rightarrow \textrm{coefficient of dynamic friction}\\N\rightarrow \textrm{Normal force}\\N=mg\\\therefore f=\mu_dmg$

Now, from Newton's second law, net force is equal to the product of mass and acceleration.

Therefore,

$\mu_dmg=ma\\\mu_d=\frac{a}{g}$

Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,

$\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025$

Therefore, the coefficient of dynamic friction is 0.025.

3 0

3 years ago

A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at

Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

Flux through spherical surface = charge inside sphere / ε₀

= 0 / ε₀

= 0 Ans .

3 0

3 years ago

A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0

algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats $m_1$ and $m_2$ , and the length of the rod $L$ .

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$ ,

solving for $v$ , we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for $m_1$ , $m_2$ , $g$ , and $L$ we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}}$

$\boxed{ v= 1.64m/s}$

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0

2 years ago

49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner

faust18 [17]

Answer:

$\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}$

Explanation:

For this interesting problem, we use the definition of centripetal acceleration

a = v² / r

angular and linear velocity are related

v = w r

we substitute

a = w² r

the rectangular body rotates at an angular velocity w

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A

the distance OB = L₂

the distance AB = L₁

the sides of the rectangle

It is indicated that the acceleration in in A and B are related

$a_A = n \ a_B$

we substitute the value of the acceleration

w² r_A = n r_B

the distance from the each corner is

r_B = L₂

r_A = $\sqrt{L_1^2 + L_2^2}$

we substitute

\sqrt{L_1^2 + L_2^2} = n L₂

L₁² + L₂² = n² L₂²

L₁² = (n²-1) L₂²

4 0

3 years ago