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Flura [38]
3 years ago
11

How many covalent bonds can be formed by Nitrogen? Please explain

Physics
1 answer:
natka813 [3]3 years ago
6 0
Three bonds

<span>Here are the combinations of hydrogen with boron, carbon, nitrogen, oxygen and fluorine. Generally to complete an octet Group 4A will form four bonds, Group 5A will form 3 bonds, Group 6A will form 2 bonds, etc. Three bonds: Boron is in group 3A. It has three valence electrons.</span>
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What does the term “retroperitoneal” mean in the terms of the location of the kidneys ?
lakkis [162]

Answer:

"Retroperitoneal" refers to the back of the peritoneum, the membrane that lines the anatomical space in the abdominal cavity. Kidney stones may cause pain to the organs within the retroperitoneal space. A diagram of the aorta, a retroperitoneal structure.

Explanation:

5 0
3 years ago
A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its centre and perpendicula
astraxan [27]

Complete Question:

A uniform rod of mass 1.90 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.40 kgis  attached to one end and a second mass m2 = 2.50 kg is attached to the other end of the rod. Treat the two masses as point particles.

(a) What is the moment of inertia of the system?

(b) If the rod rotates with an angular speed of 2.70 rad/s, how much kinetic energy does the system have?

(c) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?

(d) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.70 rad/s?

Answer:

a) 8.53 kg*m² b) 31.1 J c) 7.9 kg*m² d) 28.8 J

Explanation:

a) If we treat to the two masses as point particles, the rotational inertia of each mass will be the product of the mass times the square of the distance to the axis of rotation, which is exactly the half of the length of the rod.

As the mass has not negligible mass, we need to add the rotational inertia of the rod regarding an axis passing through its centre, and perpendicular to its length.

The total rotational inertia will be as follows:

I = M*L²/12 + m₁*r₁² + m₂*r₂²

⇒ I =( 1.9kg*(2.00)²m²/12) + 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

⇒ I =  8.53 kg*m²

b)  The rotational kinetic energy of the rigid body composed by the rod and  the point masses m₁ and m₂, can be expressed as follows:

Krot = 1/2*I*ω²

if ω= 2.70 rad/sec, and I = 8.53 kg*m², we can calculate Krot as follows:

Krot = 1/2*(8.53 kg*m²)*(2.70)²(rad/sec)²

⇒ Krot = 31.1 J

c) If the mass of the rod is negligible, we can remove its influence of the rotational inertia, as follows:

I = m₁*r₁² + m₂*r₂² = 5.40 kg*(1.00)²m² + 2.50 kg*(1.00)m²

I = 7.90 kg*m²

d) The new rotational kinetic energy will be as follows:

Krot = 1/2*I*ω² = 1/2*(7.9 kg*m²)*(2.70)²(rad/sec)²

Krot= 28.8 J

7 0
3 years ago
Una ave vuela a una velocidad constante de 15m/s en una trayectoria rectilínea. Si dura una hora volando ¿cuanta distancia habrá
earnstyle [38]

Answer:

54,000

Explanation:

3 0
3 years ago
Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an
Paladinen [302]

Answer:

\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F=m \times a

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

  • m= 8 kg
  • a= 1.6 m/s²

Substitute these values into the formula.

F= 8 \ kg * 1.6 \ m/s^2

Multiply.

F= 12.8 \ kg*m/s^2

<h3>2. Acceleration of the 2nd Object </h3>

Now,  use the force we just calculated to complete the second part of the problem. We use the same formula:

F= m \times a

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

  • F= 12.8 kg *m/s²
  • m= 2 kg

Substitute the values into the formula.

12.8 \ kg*m/s^2= 2 \ kg *a

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}

\frac {12.8 \ kg*m/s^2}{2 \ kg}=a

The units of kilograms cancel.

\frac {12.8}{2}\ m/s^2=a

6.4 \ m/s^2=a

The acceleration is 6.4 meters per square second.

4 0
3 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
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