A rubber band is launched horizontally by a student. If the air resistance is negligible, which statement BEST

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0

3 years ago

Calculate the change in the energy of an electron that moves from the n = 3 level to the n = 2 level. What type of light is emit

marissa [1.9K]

Answer:

Red light

Explanation:

The energy emitted during an electron transition in an atom of hydrogen is given by

$E=E_0 (\frac{1}{n_2^2}-\frac{1}{n_1^2})$

where

$E_0 = 13.6 eV$ is the energy of the lowest level

n1 and n2 are the numbers corresponding to the two levels

Here we have

n1 = 3

n2 = 2

So the energy of the emitted photon is

$E=(13.6) (\frac{1}{2^2}-\frac{1}{3^2})=1.9 eV$

Converting into Joules,

$E=(1.9 eV)(1.6\cdot 10^{-19} J/eV)=3.0\cdot 10^{-19} J$

And now we can find the wavelength of the emitted photon by using the equation

$E=\frac{hc}{\lambda}$

where h is the Planck constant and c is the speed of light. Solving for $\lambda$ ,

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{3.0\cdot 10^{-19}}=6.63\cdot 10^{-7} m = 663 nm$

And this wavelength corresponds to red light.

5 0

2 years ago

A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile

solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

$d=28in\\r=d/2=28/2=14in\\v=50mi/hr$

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

as

1 mile=63360 inches

1 hour=60 minutes

so

$v=(50*\frac{63360}{60} )\\v=52800in/min$

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

$w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min$

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

$N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM$

3 0

2 years ago

A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2

andrey2020 [161]

Work needed = 23,520 J

<h3>Further explanation</h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object

Work is the product of force with the displacement of objects.

Can be formulated

W = F x d

W = Work, J, Nm

F = Force, N

d = distance, m

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

5 0

3 years ago

I AM........ INEVITABLE

Archy [21]

Answer:

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5 0

3 years ago