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disa [49]
2 years ago
6

Hey guys, I don't know what this is. Any help?

Physics
1 answer:
Marizza181 [45]2 years ago
4 0
Real and reduced is the answer
You might be interested in
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
What is kepler's law??​
Elan Coil [88]
<h2>QUESTION:- </h2>

➜what is kepler's law??

\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}

Kepler gave the three laws or theorems of motion of the orbitals bodies

{\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

{\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

{\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

{T}^{2} = {a}^{3}T2=a3

where:-

T = time of revolution

a is the distance between the planet and star.

\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}

8 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
Consider the addition of an electron to the following atoms from the fourth period. Rank the atoms in order from the most negati
Murrr4er [49]
Electron configurations:

Ge: [Ar] 3d10 4s2 4p2 => 6 electrons in the outer shell

Br: [Ar] 3d10 4s2 4p5 => 7 electrons in the outer shell

Kr: [Ar] 3d10 4s2 4p6 => 8 electrons in the outer shell

The electron affinity or propension to attract electrons is given by the electronic configuration. Remember that the most stable configuration is that were the last shell is full, i.e. it has 8 electrons.

The closer an atom is to reach the 8 electrons in the outer shell the bigger the electron affinity.

Of the three elements, Br needs only 1 electron to have 8 electrons in the outer shell, so it has the biggest electron affinity (the least negative).

Ge: needs 2 electrons to have 8 electrons in the outer shell, so it has a smaller (more negative) electron affinity than Br.

Kr, which is a noble gas, has 8 electrons and is not willing to attract more electrons at all, the it has the lowest (more negative) electron affinity of all three to the extension that really the ion is so unstable that it does not make sense to talk about a number for the electron affinity of this atom.




3 0
2 years ago
A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s
TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

vf^{2} =vo^{2}+2*g*h\\

We clear the vo (initial speed)

vo=\sqrt{vf^{2}-2*g*h }

v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}

vo=5.87m/s

7 0
2 years ago
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