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3 years ago

Why does atomic size increase as you go down a grouping in the table

Physics

1 answer:

sp2606 [1]3 years ago

4 0

Its when we go down the group not up because the no. of electrons increases which results in increased no. of shells.

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An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0

3 years ago

A spring has a constant k= 10.32 N/m and is hung vertically. A 0.400kg mass is suspended from the spring. What is the displaceme

alukav5142 [94]

Answer:

displacement (x) = 0.003798 meters

Explanation:

from the fact that the string is hung vertically we can deduce that:

Total force acting on the mass = Fs (by spring) + Fg (by gravity)

where

Fs = k*x , x is the displacement..

Fg = m*g

then:

Ftot = m*a, but a = 0 m/(s^2) because the mass becames stationary.

Ftot = 0

Fs + Fg = 0

by direction, take down as negative.

Fs - Fg = 0

k*x = m*g

x = m*g/k = [(0.400)(9.8)]/(10.32)

= 0.3798 meters

4 0

3 years ago

A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c

icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

$W = F x = \mu N x = \mu\ m\ g\ x$

$W = 150 \times 6$

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

$\mu = \frac {F}{m\timesg}$

$= \frac{150}{40\times 9.8}$

= .383

We simply applied the above formulas so that each one part could calculate

3 0

4 years ago

What is atmospheric pressure ? How can we calculate it?

oksian1 [2.3K]

Answer:

Atmospheric pressure is the pressure exerted by the weight of the atmosphere.

Atmospheric Pressure = density of Mercury X acceleration due to gravity X height of column of the mercury

3 0

3 years ago

Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f

ruslelena [56]

Answer:

Moc = -613.25 [lb*in]

Explanation:

Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.

Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.

$M_{o}=r_{A/O} * F_{AB}$ (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

A (12,0,0) [in]

B (0, 24,8) [in]

C (12,24,0) [in]

$r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k$

El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].

$F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]$

De esta manera realizando el producto cruz tenemos

$M_{O}=r_{A/O} * F_{AB}$

$M_{O}=0i-685.7j+2057.1k [Lb*in]$

Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.

$OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j +0k$

El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC

$M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]$

7 0

3 years ago