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3 years ago

How is Bohr's atomic model different from Rutherford’s model?

2 answers:

8 0

In Rutherford's model, electrons were placed randomly around the nucleus. In Bohr's model, the electrons are placed in specific energy levels.

astra-53 [7]3 years ago

8 0

Answer:

Explanation:

According to the Rutherford's atomic model, all the electrons are revolving around the nucleus and experiences a centripetal force. Due to which they loses energy continuously.

Atom should emit a continuous spectrum.

According to Bohr's atom model, the electrons can revolve in the orbits having angular momentum is equal to the integral multiple of h/2π. They release energy only if they jump from higher level to the lower level.

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Anybody help me? my H.W. Q 14-17 thank you, please show work

Kipish [7]

14. Reactants:

Magnesium Chloride: $Mg Cl_2$ , since magnesium (Mg) has 2 valence electrons while chlorine (Cl) has 1 vacancy, so they are in ratio 1:2

Sodium oxide: $Na_2 O$ , since sodium (Na) has 1 valence electron while oxygen (O) has 2 vacancies, so they are in ratio 2:1

15. Products:

Magnesium oxide: $Mg O$ , since magnesium (Mg) has 2 valence electrons and oxygen (O) has 2 vacancies, so they are in ratio 1:1

Sodium chloride: $NaCl$ , since sodium (Na) has 1 valence electron and chlorine (Cl) has 1 vacancy, so they are in ratio 1:1

16: Balanced reaction:

$Mg Cl_2 + Na_2 O \rightarrow Mg O+2NaCl$

we see that on the reactant side we have: 1 atom of Mg, 2 atoms of Cl, 2 atoms of Na and 1 atom of O. By putting a 2 in front of NaCl, we have exactly the same on the product side: 1 atom of Mg, 2 atoms of Cl, 2 atoms of Na and 1 atom of O.

17. It is a double-replacement reaction

A double-replacement reaction occurs when two ionic compounds exchange their ions. In this specific case, we see that the Chlorine (initially bond with the magnesium) is then bond with the sodium, while the oxygen (initially bond with the sodium) after the reaction is bond with the magnesium.

4 0

3 years ago

small block with a mass of 0.150 kg is attached to a cord passing through a hole in a frictionless, horizontal surface. The bloc

Gre4nikov [31]

Answer:

a. Thus, the tension in the cord causes no change in angular momentum.

b. 0.112 N

c. -0.048 J

Explanation:

a) Explain if the tension force causes the change of angular momentum.

A centripetal force F initially acts on the block to keep at at a radius of 0.5 m and speed of 0.9 m/s.

When the cord is pulled, the tension causes a change in the centripetal force to a new value at a radius of 0.23 m.

But since the force is applied in the radial direction and not perpendicular to the radius, there is no torque applied (since torque τ = rFsinθ and θ = 0 thus τ = rFsin0 = 0)and thus, there is tangential acceleration α (since α = τ/I where I = rotational inertia of block)and thus no change in angular speed ω (since α = Δω/Δt, α = 0 ⇒ Δω/Δt ⇒ Δω = 0).

Since there is a no change in angular speed, there is thus no change in angular momentum.

Thus, the tension in the cord causes no change in angular momentum.

b) What is the tension in the cord in the final situation when the block moving in a circle with smaller radius.

The centripetal force in the cord is equal to the tension at the smaller radius.

We find the angular speed of the block from v = rω where v = initial tangential speed of block at r = 0.5 m = 0.9 m/s and r = 0.5 m.

So, ω = v/r = 0.9 m/s ÷ 0.5 m = 1.8 rad/s

The centripetal force at radius r = 0.23 m is F = mrω² where m = mass of block = 0.150 kg, r = distance = 0.23 m and ω = angular speed = 1.8 rad/s

So, F = 0.150 kg × 0.23 m × (1.8 rad/s)²

F = 0.150 kg × 0.23 m × 3.24 rad²/s²

F = 0.11178 kgmrad²/s²

F ≅ 0.112 N

c) How much work was done by the person who pulled on the cord

From work-kinetic energy principles, the work done by the tension equals the kinetic energy change of the block.

ΔK = W

1/2m(v₂² - v₁²) = W where m = mass of block = 0.150 kg, v₁ = initial speed of block = 0.9 m/s, v₂ = final speed of block distance r = 0.23 m,

Since the tangential speed v ∝ r the radial distance,

v₂/v₁ = r₂/r₁

v₂ = (r₂/r₁)v₁

= 0.23 m × 0.9 m/s ÷ 0.5m

= 0.207 m²/s ÷ 0.5 m

= 0.414 m/s

So, W = 1/2m(v₂² - v₁²)

W = 1/2 × 0.150 kg((0.414 m/s)² - (0.9 m/s)²)

W = 1/2 × 0.150 kg((0.1714 m/s)² - (0.81 m/s)²)

W = 1/2 × 0.150 kg((-0.6386 m²/s²)

W = 1/2 × -0.09579 kgm²/s²)

W = -0.0479 kgm²/s²

W = -0.0479 J

W ≅ -0.048 J

3 0

3 years ago

Two billiard balls with the same mass undergo a perfectly elastic head-on collision. if one ball's initial speed was 2 m/s, and

My name is Ann [436]

The ball with an initial velocity of 2 m/s rebounds at 3.6 m/s
The ball with an initial velocity of 3.6 m/s rebounds at 2 m/s

There are two principles involved here
Conservation of momentum and conservation of energy.
I'll use the following variables
a0, a1 = velocity of ball a (before and after collision)
b0, b1 = velocity of ball b (before and after collision)
m = mass of each ball.

For conservation of momentum, we can create this equation:
m*a0 + m*b0 = m*a1 + m*b1
divide both sides by m and we get:
a0 + b0 = a1 + b1

For conservation of energy, we can create this equation:
0.5m(a0)^2 + 0.5m(b0)^2 = 0.5m(a1)^2 + 0.5m(b1)^2
Once again, divide both sides by 0.5m to simplify
a0^2 + b0^2 = a1^2 + b1^2

Now let's get rid of a0 and b0 by assigned their initial values. a0 will be 2, and b0 will be -3.6 since it's moving in the opposite direction.
a0 + b0 = a1 + b1
2 - 3.6 = a1 + b1
-1.6 = a1 + b1
a1 + b1 = -1.6

a0^2 + b0^2 = a1^2 + b1^2
2^2 + -3.6^2 = a1^2 + b1^2
4 + 12.96 = a1^2 + b1^2
16.96 = a1^2 + b1^2
a1^2 + b1^2 = 16.96

The equation a1^2 + b1^2 = 16.96 describes a circle centered at the origin with a radius of sqrt(16.96). The equation a1 + b1 = -1.6 describes a line with slope -1 that intersects the circle at two points. Those points being (a1,b1) = (-3.6, 2) or (2, -3.6). This is not a surprise given the conservation of energy and momentum. We can't use the solution of (2, -3.6) since those were the initial values and that would imply the 2 billiard balls passing through each other which is physically impossible. So the correct solution is (-3.6, 2) which indicates that the ball going 2 m/s initially rebounds in the opposite direction at 3.6 m/s and the ball originally going 3.6 m/s rebounds in the opposite direction at 2 m/s.

6 0

4 years ago

Radio waves and microwaves are similar because they both have

tekilochka [14]

The correct answer is

b. low frequencies and low energy.

In fact, microwaves and radio waves have frequency up to $10^{10}-10^{11} Hz$ , which are really low compared, for instance, to the frequency of gamma rays ( $10^{20} Hz$ ). Moreover, the energy of an electromagnetic wave is directly proportional to its frequency:

$E=hf$

where h is the Planck constant and f the frequency; therefore, low frequency means also low energy.

4 0

3 years ago

10 poin What is the momentum of a 25kg cart traveling left with a velocity of 10 m/s?

Mashutka [201]

250 kg*m/s
Momentum is found by multiplying mass by velocity
25*10=250

7 0

3 years ago