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Natalka [10]
2 years ago
3

A single electron has a charge of 2.9×10−15 ZZ . Using zeets and a fictitious unit of mass called the wiggle, ww, the charge-to-

mass ratio of an electron is 2.4×106 Z/wZ/w . What is the mass of an electron in wiggles? Express the mass of an electron in wiggles using two significant figures.
Physics
1 answer:
fenix001 [56]2 years ago
7 0

Answer:

m = 1.2 10⁻²¹  w

Explanation:

Despite having other units the laws of physics and comply with all inertial systems, so the charge-mass ratio of the electron is the same

                         Rel = e / m

                         m = e / Rel

Where Rel is the ratio, e the charge of the electron and m the masses of the electron

Let's calculate

                        m = 2.9 10⁻¹⁵ / 2.4 10⁶

                       m = 1.2 10⁻²¹  w

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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
Read 2 more answers
Which of the following statements characterizing types of waves are true?
Tasya [4]

Answer:

a and b.

Explanation:

In general  types of wave

1. Transverse wave  :

    In these waves particle are vibrate perpendicular to motion of waves.

 Ex : Electromagnetic wave , Radio wave .

2. Longitudinal wave :

   In these waves particle are vibrate along the motion of waves.

 Ex : Sound wave

Mechanical wave  :

1 .These are transverse wave or Longitudinal wave or combination of them .

2.These waves required medium for propagation.

3. The particle are vibrate perpendicular to motion of waves.

So the option a and b are correct.

7 0
2 years ago
Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom
Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

B = B_o(-\hat k)

now the velocity is from bottom to top

so we have

v = v_o \hat j

now the force on the moving charge is given as

\vec F = q(\vec v \times \vec B)

now we have

\vec F = (-e)(v_o \hat j \times B_o(-\hat k))

\vec F = e v_o B \hat i

so force will be towards Right

7 0
2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
Strike441 [17]

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer f_{obs} is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

f_{obs} = f_s (\frac{v_w}{v_w-v_s})

f_s= Frequency of the source

v_w= Speed of sound

v_s= Speed of source

The velocity of the ambulance is

v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

v_s = 30.55m/s

Replacing at the expression to frequency of observer we have,

f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})

f_{obs} = 878Hz

Therefore the frequency receive by observer is 878Hz

8 0
2 years ago
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