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3 years ago

Compare the characteristics of 4d orbitals and 3d orbitals and compared the following sentences. Check all that apply

Physics

1 answer:

charle [14.2K]3 years ago

8 0

Answer :

<em>(b) 4d orbitals would be larger in size than 3d orbitals</em>

<em>(e) 4d orbitals would have more nodes than 3d orbitals</em>

Explanation :

As we move away from one orbital to another, the distance between nucleus and orbital increases. So, 4d orbitals would be far to the nucleus than 3d orbitals.

Hence, 4d orbitals would be larger in size than 3d orbitals.

Number of nodes is any orbital is n - 1 where, n is principal quantum number.

So, number of orbital in 4d is 3.

And number of orbital in 3d is 2.

So, options (b) and (e) are correct.

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Your film idea is about drones that take over the world. In the script, two drones are flying horizontally at the same speed and

Stella [2.4K]

Answer:

vₓ = 20 m/s, v_{y} = -15 m / s

Explanation:

This is a conservation of moment problem, since it is a vector quantity we can work each axis independently

The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved

X axis

Initial moment. Before the crash

p₀ = m₁ v₀ₓ + m₂ v₀ₓ

Final moment. After the crash

p_{fx} = (m₁ + m₂) vₓ

p₀ₓ = $p_{fx}$

m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ

vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)

vₓ = v₀ₓ = 20 m/s

Y Axis

Initial

p_{oy} = m₁ v_{oy}

Final

p_{fy} = (m₁ + m₂) v_{y}

p_{oy} = p_{fy}

the drom rises and when it falls it has the same speed because there is no friction v_{oy} = -60 m/s

m₁ $v_{oy}$ = (m₁ + m₂) v_{y}

v_{y} = m₁ / (m₁ + m₂) v_{oy}

v_{y} = 1/4 60

v_{y} = -15 m / s

Vertical speed is down

5 0

4 years ago

a ball is dropped from rest at a height of 45.0 m above the ground. ignore the effects of air resistance. What is the speed of c

NemiM [27]

So, the final velocity of the ball when it is 10.0 m above the ground approximately <u>26.2 m/s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses the principle of final velocity in free fall. Free fall occurs only when an object is dropped (without initial velocity), so the falling object is only affected by the presence of gravity. In general, the final velocity in free fall can be expressed by this equation :

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$

With the following condition :

v = final velocity (m/s)
h = height or any other displacement at vertical line (m)
g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

$\sf{h_1}$ = initial height = 45.0 m
$\sf{h_2}$ = final height = 10.0 m
g = acceleration of the gravity = 9.8 m/s²

Note :

At this point 10 m above the ground, the object can still complete its movement up to exactly 0 m above the ground.

What was asked :

v = final velocity = ... m/s

Step by Step

$\sf{v = \sqrt{2 \times g \times \Delta h}}$

$\sf{v = \sqrt{2 \times g \times (h_1 - h_2)}}$

$\sf{v = \sqrt{2 \times 9.8 \times (45 - 10)}}$

$\sf{v = \sqrt{19.6 \times 35}}$

$\sf{v = \sqrt{686}}$

$\boxed{\sf{v \approx 26.2 \: m/s}}$

<h3>Conclusion</h3>

So, the final velocity of the ball when it is 10.0 m above the ground approximately 26.2 m/s.

The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625

5 0

2 years ago

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont

koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216

7 0

2 years ago

A parallel plate capacitor above has square plates 10 cm across on each side. The plates are 50 μm apart, and the electric field

IgorLugansk [536]

Answer:

a) ΔV = 20 V , b) Q = 35.4 10⁻⁹ C, c) C = 1.77 10⁻⁹ F

Explanation:

a) The electric potential is

ΔV = E x

ΔV = 4.0 10⁵ 50 10⁻⁶

ΔV = 20 V

c) The capacity of a capacitor is

C = ε₀ A / d

The area of a square plate is

A = L²

A = 0.10²

A = 0.01 m²

L = 50mm = 50 10⁻⁶ m

Let's calculate

C = 8.85 10⁻¹² 0.01 /50 10⁻⁶

C = 1.77 10⁻⁹ F

b) the charge is

Q = C ΔV

Q = 1.77 10⁻⁹ 20

Q = 35.4 10⁻⁹ C

4 0

3 years ago

Read 2 more answers

An object starts from rest and accelerates to a velocity of 10 m/s, east in 5 seconds. What is the acceleration of the object?

Shalnov [3]

Answer:

2m/s²

Explanation:

acceleration is the rate of change of velocity

for velocity to reach 10m/s in 5s,

acceleration= 10/5

3 0

4 years ago