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zalisa [80]
3 years ago
12

What type of seismic waves travel through Earth?

Physics
2 answers:
ycow [4]3 years ago
8 0

Answer:d. Ground waves

Explanation:A P-wave is one of the two main types of seismic waves. 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

-Dominant- [34]3 years ago
7 0

<u>Answer:</u>

The correct options are A and C.

<u>Explanation :</u>

When there is a sudden release of energy due to earthquake or any type of explosion inside the earth, then this energy is transferred in the form of waves called Seismic Waves.

Seismograph helps us to measure these waves. These seismic waves can be categorized as Body Waves and Surface Wave.  

Body Waves propagate inside the earth in all directions, and are further classified as Primary Waves (P-Waves) and Secondary Waves (S-Waves).The Surface Waves move along the earth’s surface, like the ripples on water surface.

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Find the mass if the force is 18 N and the acceleration is 2 m/s2.
Flura [38]

Answer:

<h3>The answer is 9 kg</h3>

Explanation:

To find the mass given the force and acceleration we use the formula

m =  \frac{f}{a}  \\

where

f is the force

a is the acceleration

We have

f = 18 N

a = 2 m/s²

We have

m =  \frac{18}{2}  \\

We have the final answer as

<h3>9 kg</h3>

Hope this helps you

4 0
3 years ago
A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

E=\phi + K (1)

where

E=\frac{hc}{\lambda} is the energy of the incident photon, with

h = Planck constant

c = speed of light

\lambda = wavelength

\phi = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

eV=K

So we can rewrite (1) as

E=\phi + eV

where we have:

\lambda=200 nm = 2\cdot 10^{-7} m

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J

Converting into electronvolts,

E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV

And now we can solve eq.(1) to find the work function of the metal:

\phi = E-eV=6.22 eV-1.93 eV=4.29 eV

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so \phi is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0
3 years ago
Please help me, I’m confused on where to start.
zimovet [89]

Answer:

D

Explanation:

6 0
3 years ago
Read 2 more answers
What is the resistance (R) when voltage is 179V and current is 5 Amps?
Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

3 0
3 years ago
Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i
densk [106]

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

6 0
3 years ago
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