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2 years ago

What is common in elastic collision, inelastic collision and perfectly inelastic collision

Physics

2 answers:

emmainna [20.7K]2 years ago

8 0

Answer:

In the all types of collision, momentum is conserved.

Explanation:

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Feliz [49]2 years ago

7 0

Answer:

A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision.

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A disk of mass m and moment of inertia of I is spinning freely at 6.00 rad/s when a second identical disk, initially not spinnin

Nadusha1986 [10]

Answer:

The angular speed of the new system is $3\,\frac{rad}{s}$ .

Explanation:

Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:

$I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}$

Where:

$I$ - Moment of inertia of a disk, measured in kilogram-square meter.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\omega_{f}$ - Final angular speed, measured in radians per second.

This relationship is simplified and final angular speed can be determined in terms of initial angular speed:

$\omega_{f} = \frac{1}{2}\cdot \omega_{o}$

Given that $\omega_{o} = 6\,\frac{rad}{s}$ , the angular speed of the new system is:

$\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)$

$\omega_{f} = 3\,\frac{rad}{s}$

The angular speed of the new system is $3\,\frac{rad}{s}$ .

6 0

2 years ago

The standard wave format for any wave is wave. When depicting wave in standard wave format, the direction of motion must be rota

german

Answer:

Transverse wave and Longitudinal wave and Electromagnetic wave

Explanation:

An inverted wave is a wave in which the vibrations of the particles are perpendicular to the direction of wave motion.
Longitudinal waves, on the other hand, are waves in which the vibrations of the particles are parallel to the direction of wave motion.
Electromagnetic waves are waves that do not require medium media for transmission, including radio waves, microwaves, UV lights, etc.
Most electromagnetic waves are transverse in nature.

4 0

2 years ago

A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density

Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

mole fraction (H2SO4)= 0.034
mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

n = m / MW
= ∑ AWT
mole fraction = moles of A component / total moles of solution
ρ = m /v

Where:

m = mass
n = moles
MW = molecular weight
AWT = atomic weight
ρ = density
v = volume

Information about the problem:

m solute (H2SO4) = 17.75 g
v(solution) = 100 ml
ρ (solution)= 1.094 g/ml
AWT (H)= 1 g/mol
AWT (S) = 32 g/mol
AWT (O)= 16 g/mol
mole fraction(H2SO4) = ?
mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)= 98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)= 18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092 mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092 mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092 mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

Solvent
Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

8 0

1 year ago

A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : $f=20\ cm$

Height of object : $h=1\ cm$

Object distabce from lens : $u=-10\ cm$

Using lens formula: $\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ , we get

$\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}$ , where v = image distance from the lens.

On solving aboive equation , we get

$\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm$

Formula of Magnification : $m=\dfrac{v}{u}=\dfrac{h'}{h}$ , where h' is the height of image.

Put value of u, v and h in it , we get

$\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm$

Hence, the height of the image is 2 cm.

3 0

2 years ago

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti

allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q = −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0

3 years ago

What is common in elastic collision, inelastic collision and perfectly inelastic collision​

What is common in elastic collision, inelastic collision and perfectly inelastic collision