Which best describes a neutralization reaction?

An electron and a second particle both move in circles perpendicular to a uniformmagnetic field. The mass of the second particle

Katarina [22]

Answer:

The change on the second particle is $2.93\times 10^{-16}\ C$ .

Explanation:

The period of revolution of the particle in the magnetic field is given by the formula as follows :

$T=\dfrac{2\pi m}{Bq}$

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

$m_s=m_p$

If both particles take the same amount of time to go once around their respective circles. So,

$T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C$

So, the change on the second particle is $2.93\times 10^{-16}\ C$ .

7 0

3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

So

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

3 0

2 years ago

A Universe Question!

vfiekz [6]

I'm pretty sure this is A

4 0

3 years ago

Todo sistema heterogêneo é bifásico

strojnjashka [21]

I don't understand the language.....

3 0

3 years ago

A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck The puck travels for

Norma-Jean [14]

Answer:

Explanation:

Let's analyze the situation presented in order to know which answer is correct.

When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.

Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.

The correct answer is:

C) Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck

4 0

2 years ago