If the valence electrons were removed, what would be the ion charge of the element?

While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it

diamong [38]

Answer: 6s

Explanation:

Vs=32m/s speed at beginning of slowing down

Vf=0m/s stop speed

a= -6 m/s² acceleration

----------------

Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

t=6 s

7 0

3 years ago

2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after

sweet [91]

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

$A(t)=A_oe^{\frac{-bt}{2m}}$

or amplitude for unforce oscillation

$\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}$

or

$\frac{6bt}{2m}=1$

or

$b=\frac{m}{3T}$

Now, provided in the question Amplitude of the driven oscillation

$A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}$

the value of the maximum amplitude is obtained $(k=m\omega_d^2)$

thus,

$A=\frac{F_{max}}{(b\omega_d}$

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

$0.075=\frac{nF}{((\frac{m}{3T})2\pi}$

or

n = 1810

hence, there were 1810 people on the bridge

b) $A=\frac{F_{max}}{(b\omega_d}$

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

$A=\frac{F_{max}}{(3b\omega_d}$

or

$A=\frac{1}{(3}A_o$

or

$A=\frac{1}{(3}0.075$

or

A = 0.025 m = 25 mm

6 0

3 years ago

1:What kind of energy does your body use to keep you warm.

Zigmanuir [339]

Answer:

1. thermal energy

2. Batteries and coal

4. energy conversion

3 0

3 years ago

A 2.5 m segment of wire supplying current to the motor of a submerged submarine carries 1050 A and feels a 3.6 N repulsive force

Zanzabum

force between two parallel wire is

f/l = mueo*i1i2/2pir

f/l = 2*10^-7*i1i2/rl

i2 = f*r/2*10^-7*i1

i2 = 342.9 A

8 0

3 years ago

harge of uniform density (100 nC/m3) is distributed throughout a hollow cylindrical region formed by two coaxial cylindrical sur

Elden [556K]

Answer:

The magnitude of the electric field is 8.47 N/C

Explanation:

Given;

uniform charge density, λ = 100 nC/m³

inner radii of the cylinder, r = 1.0 mm and 3.0 mm

distance from the symmetry axis, R = 2.0 mm

$Volume =\pi (R^2 -r^2)l\\\\Volume =\pi ((2*10^{-3})^2 -(1*10^{-3})^2)l\\\\Volume =\pi (4*10^{-6} - 1*10^{-6})l\\\\Volume = 3*10^{-6} \pi l \ m^3$

Area = 2πrl

Area =2π(2 x 10⁻)l

Volume = A x d

d = Volume / Area

$d = \frac{V}{A} = \frac{3*10^{-6}*\pi*l}{4\pi *10^{-3} l} = 75 *10^{-5} \ m$

the magnitude of the electric field

$E = \frac{\lambda *d}{\epsilon_o} = \frac{100*10^{-9} *75*10^{-5}}{8.854*10^{-12}} \\\\E = 8.47 \ N/C$

Therefore, the magnitude of the electric field is 8.47 N/C

6 0

3 years ago