Which of the following statements is not true of vibrations?

Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen

Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

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5 0

2 years ago

A skateboarder is moving 5.25 m/s when he starts to roll up a frictionless hill. How much higher is he when his velocity has red

Taya2010 [7]

Mechanical energy E = mgh + 1/2mv²

When he starts, let h = 0 ⇒ E₁ = 1/2mv₁²
When he reaches height h ⇒ E₂ = mgh + 1/2mv₂²

Without friction, energy is conserved at all times.

E₁ = E₂
↓
1/2mv₁² = mgh + 1/2mv₂²
↓
1/2v₁² = gh + 1/2v₂²
↓
gh = 1/2(v₁² - v₂²)
↓
h = (v₁² - v₂²) / (2g)

5 0

4 years ago

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For each star, determine how its light would be shifted. Not all choices may be used, and some may be used more than once. A red

barxatty [35]

Answer:

Explanation:

To calculate the red shift you use the following formula:

$z=\frac{1+vcos\theta/c}{\sqrt{1-v^2/c^2}}-1$

\tetha: angle between the observer and the motion of the body

v: speed of the body

c: speed of light

for motion with angle 90° (transversal motion):

$z=\sqrt{\frac{c+v}{c-v}}-1$

- A red dwarf moving away from Earth at 39.1 km/s :

$z=\sqrt{\frac{3*10^8m/s+39.1*10^3m/s}{3*10^8m/s-39.1*10^3m/s}}-1=1.3*10^{-4}$

- A yellow dwarf moving transversely at 15.1 km/s (angle = 90°):

$z=\frac{1+0}{\sqrt{1-(15.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.27*10^{-9}$

- A red giant moving towards Earth at 23.3 km/s (angle = 0°):

$z=\frac{1+(23.3*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(23.3*10^3m/s)^2/(3*10^8m/s)^2}}-1=7.76*10^{-5}$

- A blue dwarf moving away from Earth at 25.9 km/s $z=\frac{1+(25.9*10^3m/s)/(3*10^8m/s)}{\sqrt{1-(25.9*10^3m/s)^2/(3*10^8m/s)^2}}-1=8.63*10^{-5}$

- A red dwarf moving transversely at 14.1 km/s

$z=\frac{1+0}{\sqrt{1-(14.1*10^3m/s)^2/(3*10^8m/s)^2}}-1=1.11*10^{-9}$

6 0

4 years ago

PLEASE HELP WILL GIVE 10 POINTS!!!!

serg [7]

D. freezing, deposition, and condensation

Explanation:

In order to understand, let's see the definition of these changes of state:

- freezing: freezing occurs when a substance changes from liquid state to solid state

- deposition: deposition occurs when a substance changes from gas state to solid state, without passing through the liquid state

- condensation: this process occurs when a substance changes from gas state to liquid state

In all the three processes listed above, energy is released from the substance, which in fact loses energy. In fact:

- in a liquid, the particles have more energy than in a solid (because in liquid they can move, while in solid they can only vibrate), so during freezing, since a liquid turns into a solid, it releases energy to the surrounding

- in a gas, the particles have more energy than in a liquid (because in a gas they can move faster than in a liquid), so during condensation, since a gas turns into a liquid, energy is released to the surrounding

- in deposition, the particles change from the gas state (where they move faster) to a solid state (where they are not free to move), so they lose energy as well.

4 0

4 years ago

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The push up is dynamic or static

77julia77 [94]

Answer:

Dynamic exercises

Explanation:

7 0

3 years ago