All categories

3 years ago

Built an atom of Krypton which has an atomic number of 36?

Physics

1 answer:

gogolik [260]3 years ago

7 0

Answer:

what's the question exactly because what you said was true but what's the question

You might be interested in

What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An

GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}$

$E=\frac{2m_{0}c^}2}{\sqrt{3}}$

$E=\frac{2\times0.001\times9\times10^{16}}{1.732}$

$E=1.04\times10^{14}J$

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, $E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV$

4 0

3 years ago

A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a

AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

$q = -Kb\frac{dh}{dl}$

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25 m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

$V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day$

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

$= 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours$

4 0

3 years ago

Irina rode her bike to work at an average speed of 16 miles per hour. It started to rain, so she got a ride home along the same

KengaRu [80]

Answer:

0.7 hours

Explanation:

From the way back, we can calculate the distance between Irina's work and Irina's home. In fact, we know that the car takes 0.4 hourse traveling at 27 mph, so the distance covered should be

$d=vt=(27 mph)(0.4 h)=10.8 mi$

When Irina rides to work with her bike, she travels at a speed of 16 mph. So we can find the time she takes by dividing the total distance (10.8 miles) by her speed:

$t=\frac{d}{v}=\frac{10.8 mi}{16 mph}=0.7 h$