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2 years ago

How can the atomic number of nitrogen be determined?

Physics

1 answer:

Doss [256]2 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.

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Identify areas where winds are weak

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3 years ago

Which equation best describes the law of conservation of momentum?

NeX [460]

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8 0

3 years ago

Read 2 more answers

A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

Korvikt [17]

Answer:

The strength of the electric field is $1.35\times10^{4}\ N/C$ .

Explanation:

Given that,

Speed $v= 5.05\times10^{5}\ m/s$

Time $t= 3.90\times10^{-7}\ s$

Angle = 45°

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$5.05\times10^{5}=0+a\times3.90\times10^{-7}$

$a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}$

$a=1.29\times10^{12}\ m/s^2$

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

$F= ma=qE$

$ma = qE$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}$

$E=1.35\times10^{4}\ N/C$

Hence, The strength of the electric field is $1.35\times10^{4}\ N/C$ .

3 0

3 years ago

Plz, help the rubber ball be dropped from the top of a ladder. It bounces on the same spot on the ground several times to a less

Vladimir [108]

Answer:

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Explanation:

6 0

2 years ago

A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot

Leokris [45]

Answer:

a. 32.67 rad/s² b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0

2 years ago