Which of the following body systems work together to get sells the oxygen they need to carry at the cellular respiration

Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls

Alina [70]

Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 = 8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

5 0

2 years ago

What is the maximum angular momentum Lmax that an electron with principal quantum number n = 2 can have? Express your answer in

butalik [34]

Answer:

$L_{max} = 1.414$ ℏ

Given:

Principle quantum number, n = 2

Solution:

To calculate the maximum angular momentum, $L_{max}$ , we have:

$L_{max} = \sqrt {l(1 + l)}$ (1)

where,

l = azimuthal quantum number or angular momentum quantum number

Also,

n = 1 + l

2 = 1 + l

l = 1

Now,

Using the value of l = 1 in eqn (1), we get:

$L_{max} = \sqrt {1(1 + 1)} = \sqrt 2$

$L_{max} = 1.414$ ℏ

4 0

3 years ago

Timed please hurry

vladimir1956 [14]

The answer is D. disorganized

5 0

2 years ago

Read 2 more answers

A fluid flows along the x axis with a velocity given by V = ( x / t ) ˆ i , where x is in feet and t in seconds. (a) Plot the sp

umka21 [38]

Answer:

c)

V_local = -x/t^2

V_convec = x/t^2

d)

a = V_local + V_convec = 0

e) When a particle moves towards postive x direction its convective velocity increases, but at the same time the local velocity deacreases (at the same rate) when time increases

Explanation:

Hi!

You can see plots for a) and b) attached on this document

c)

The local acceleration is just teh aprtial derivative of the velocity with respect to t:

$\frac{dV}{dt} = \frac{d}{dt} \frac{x}{t}=- \frac{x}{t^2}$

And the convective acceleration is given by the product of the velocity times the gradient of the velocity, that is:

$\vec{v} \cdot \nabla \vec{v} = v ( \frac{dv}{dx} ) =\frac{x}{t} \frac{1}{t} = \frac{x}{t^2}$

d)

Since the acceleration of any fluid particle is the sum of the local and convective accelerations, we can easily see that it is equal to zero, since they are equal but with opposit sign

e)

This is because of teh particular form of the velocity. A particle will move towards areas of higher velocities (convectice acceleration), but as time increases, the velocity is also decreasing (local acceleration), and the sum of these quantities adds up to zero