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melamori03 [73]
3 years ago
12

The type of seismic waves that arrive at the surface first and move by compressing and expanding the ground like an accordion ar

e called
a. S waves.
b. P waves.
c. Surface waves.
d. Mercalli waves.
Physics
2 answers:
balu736 [363]3 years ago
5 0

Answer:

b. P waves

Explanation:

A P-wave is one of the two main types of elastic body waves, called seismic waves in seismology. P-waves travel faster than other seismic waves and hence are the first signal from an earthquake to arrive at any affected location or at a seismograph. P-waves may be transmitted through gases, liquids, or solids. Primary waves are alternatingly compressional and extensional, and cause the rocks they pass through to change in volume. These waves are the fastest traveling seismic waves and can travel through solids, liquids, and gases. Also called P wave. See Note at earthquake. P Waves are compressional which means they move through (compress) a solid or liquid by pushing or pulling similar to the way sound travels through the air. The particles of the material a P Wave pushes through move in the direction of the P wave's energy.

LekaFEV [45]3 years ago
3 0

The answer is B. P waves.

I took the test and it was correct. I hope this helps!

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A fishbowl has a circular opening with a diameter of 13 cm. The fishbowl sits upright on a table in a magnetic field of 0.00110
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Explanation:

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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
3 years ago
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