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3 years ago

What type of wave does NOT need matter to carry energy? *

Physics

1 answer:

Blababa [14]3 years ago

3 0

Answer:

I guess sound wave I s gonna be d right answer

Explanation:

cos sound doesnt has weight and occupies space

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A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t

irinina [24]

Answer:

$M_{av}=35.521amu$

Explanation:

As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

$M_{av}=(35amu*75.8+37amu*24.3)/100=35.521amu$

4 0

2 years ago

The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid

IgorLugansk [536]

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x $x velocity^{2}$

incoming kinetic energy = 0.5 x m x $x u^{2}$

coefficient of restitution = $\frac{v}{u}$

0.72 = $\frac{v}{u}$

v = 0.72u

therefore the outgoing kinetic energy = 0.5 x m x $(0.72u)^{2}$

outgoing kinetic energy = 0.5 x m x $0.5184 x u^{2}$

outgoing kinetic energy = 0.5184 (0.5 x m x $x u^{2}$ )

recall that 0.5 x m x $x u^{2}$ is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

5 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago

PLEASE HELP ASAP!!!!!!!1

ArbitrLikvidat [17]

Age of dino: Mesozoic era

End of earth a dessert: End of the Mesozoic

a layer: Paleozoic

7 0

2 years ago

CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do

lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.

Explanation:

Given that,

Spring constant of the spring, $k=1.2\times 10^4\ N/m$

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

$E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J$

So, the elastic potential energy of the spring change during this process is 21.6 J.

4 0

2 years ago