All categories

3 years ago

How can the atomic number of nitrogen be determined?

Physics

1 answer:

Doss [256]3 years ago

4 0

Answer:

So... for the element of NITROGEN, you already know that the atomic number tells you the number of electrons. That means there are 7 electrons in a nitrogen atom. Looking at the picture, you can see there are two electrons in shell one and five in shell two.

You might be interested in

Using this formula Vj = V; + at, If a vehicle starts from rest

Furkat [3]

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

$v_f=0+4.5*8$

$v_f=36\ m/s$

The final speed of the vehicle is 36 m/s

5 0

3 years ago

The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f

Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0

3 years ago

Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0

3 years ago

Pe=___

Nesterboy [21]

Answer:

H = start height (v = 0)

h = present height

v = present speed

assuming no friction

total energy = PE + KE

mgH = mgh + .5mv^2

if PE = KE then

mgH = mgh + mgh

h = H/2

potential energy = kinetic energy when object is at half its start height.

Explanation:

8 0

3 years ago

A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t

navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that $v^2=u^2+2as$

$16^2=24^2+2\times a\times 50$

$a=-3.2m/sec^2$

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So $v^2=u^2+2as$

$0^2=24^2-2\times 3.2\times s$

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration $a=-3.2m/sec^2$

So time $t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec$

7 0

3 years ago