Body mass index (bm) is a measure of body weight relative

What number and type of hybrid orbital(s) form(s) when one p and one s atomic orbital mix?

Harrizon [31]

Two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

<h3>What are orbitals?</h3>

Orbital is the place around nucleus where mostly the electrons are present. There are four types of orbitals are present, s, p, d, and f.

The orbitals that are formed by the mixing of these orbitals are called hybrid orbitals.

Thus, two equivalent hybridized orbitals will form from the mixing of one s-orbital and one p-orbital, that is (sp) orbital.

Learn more about orbitals

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3 0

2 years ago

Which of the following is a key assumption of the scientific method

lesantik [10]

Answer:

Though you have not gave the choices, I do believe it is “testing”

Explanation:

3 0

3 years ago

A 5.00 kg pendulum swings back and forth. At the top of its arc it reaches a height of 0.36 m. What is the velocity of the pendu

azamat

Answer:

Gravitational potential energy = (mass) x (gravity) x (height)

Kinetic energy (of a moving object) = (1/2) (mass) x (speed)²

When the pendulum is at the top of its swing, its potential energy is

(mass) x (gravity) x (height)

= (5 kg) x (9.8 m/s²) x (0.36 m)

= (5 x 9.8 x 0.36) joules

17.64 joules .

Energy is conserved ... it doesn't appear or disappear ... so that number is exactly the kinetic energy the pendulum has at the bottom of the swing, only now, it's kinetic energy:

17.64 joules = (1/2) x (mass) x (speed)²

17.64 joules = (1/2) x (5 kg) x (speed)²

Divide each side by 2.5 kg:

17.64 joules / 2.5 kg = speed²

Write out the units of joules:

17.64 kg-m²/s² / 2.5 kg = speed²

(17.64 / 2.5) (m²/s²) = speed²

7.056 m²/s² = speed²

Take the square root of each side: Speed = √(7.056 m²/s²) = 2.656 m/s.

Looking through the choices, we're overjoyed to see that one if them is ' 2.7 m/s '. Surely that's IT !

Hope this will help you ...

8 0

2 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

3 years ago

A charge of +3.0 mC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energ

Marat540 [252]

Answer:

Work done = 4584.9 J

Explanation:

given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C

Solution:

Formula for the potential difference at the center of the circle

P.E = K × q1 q2 /r (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)

P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C / 0.20 m

P.E = 4584.9 J = Work done

3 0

3 years ago