How many dots belong in the electron dot diagram of a boron(B) atom

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

2 years ago

a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100

Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

let's reduce the magnitudes to the SI system

r = 10 cm = 0.10 m

m = 10 g = 0.010 kg

A = 100 cm² = 0.01 m²

the definition of density is

ρ = m / V

the volume of a sphere

V = $\frac{4}{3} \ \pi r^{3}$

V = $\frac{4}{3}$ π 0.1³

V = 4.189 10⁻³ m³

let's calculate the density of the ball

ρ = $\frac{0.010}{4.189 \ 10^{-3} }$

ρ = 2.387 kg / m³

the tabulated density of water is

ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

B - W = 0

B = W

where B is the thrust that is given by Archimedes' principle

ρ_liquid g V_submerged = m g

V_submerged = m / ρ_liquid

we calculate

V _submerged = 0.10 9.8 / 997

V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

V = A h

h = V / A

let's calculate

h = 9.83 10⁻⁴ / 0.01

h = 0.0983 m

this is equal to h = 9.83 cm

8 0

2 years ago

If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is th

Taya2010 [7]

Answer: The magnitude of the current in the second wire 2.67A

Explanation:

Here is the complete question:

Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?

Explanation: Please see the attachments below

4 0

3 years ago

A large refrigerator (mass 80kg) sits at rest inside a house. The homeowner wants to move the refrigerator across the room, so s

Liono4ka [1.6K]

Answer:

400 N

Explanation:

By the law of friction,

$F=\mu R$

$F$ is the maximum frictional force, $\mu$ is the coefficient of friction and $R$ is the reaction on the refrigerator. On a horizontal surface, the reaction is equal to the weight of the refrigerator.

$R=mg$

$F=\mu mg$

While not moving, the fricition on the refrigerator is static friction. So, $\mu=0.65$

$F=0.65 \times80\times9.8=509.6 \text{ N}$

This is the maximum frictional force and is more than the applied horizontal force of 400 N. Frictional force cannot be more than the applied force, else it would actually pull the refrigerator backwards (a strange thing, if it were to happen). It is equal to the extent of the applied force because the applied force is not enough to overcome the maximum.

Hence the frictional force is 400 N.

PS: Note that we do not use the coefficient of kinetic friction because applied force could not overcome the static friction.

6 0

3 years ago

speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?

Lina20 [59]

S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

= 1.339844m

7 0

2 years ago