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Sveta_85 [38]
3 years ago
13

Waves which require a medium are known as

Physics
1 answer:
juin [17]3 years ago
8 0
Mechanical waves require medium
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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
4 years ago
1. For safety concerning a curve, you should: A. decelerate before the curve
chubhunter [2.5K]
<span>In the question "For safety concerning a curve, you should" the correct answer is "decelerate before the curve" (option A) Decelerating before the curve gives you the opportunity to navigate the curve adequatery and be able to contain any on-coming vehicle.</span>
4 0
3 years ago
Read 2 more answers
In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They
grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

           t = 4 88

           t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

           #_rotaciones = t / to

           #_rotations = 352 / 58.7

           #_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

          t = 6 88

          t = 528 s

The number of rotations on itself is

           #_rotaciones = t / to

           #_rotations = 528 / 58.7

           #_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

           t = 8 88

           t = 704 days

the number of turns on itself is

            #_rotaciones = t / to

            #_rotations = 704 / 58.7

            #_rotations = 12

True

The only incorrect statement is from student B

6 0
3 years ago
what is the final speed of a skater who accelerates at a rate of 2.0m/s^2 from rest for 3.5s? Show your work
Advocard [28]
Okay, so lets look at the equation.

The Skate accelerates at 2.0m/s².  This can confuse some people but basically all you need to do is plug in the seconds.

So lets do 1 second.  It would be 2.0m/1² which is 2.0m.

Lets do 2 seconds.  It would be 2.0m/2² which is 0.5m.

Lets do 3 seconds.  It would be 2.0m/3² which is 0.222m

Lets do 3.5 seconds which is about 0.163m.

Add those together and you get 2.885m.

This may be wrong but this is what I got given what you gave (that its seconds squared).
3 0
3 years ago
Read 2 more answers
What does the Nucleolus do?
SCORPION-xisa [38]

Answer:

C

Explanation:

6 0
3 years ago
Read 2 more answers
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