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Sunny_sXe [5.5K]
3 years ago
14

When earthquakes produce vibrations called waves, how do those waves travel? They ripple from the epicenter of the earthquake. T

hey move back and forth and side to side through water. They travel away from the focus of the earthquake in all directions. They move slowly and increase in intensity as they reach Earth's surface.
Physics
2 answers:
Lana71 [14]3 years ago
7 0

Answer: They travel away from the focus of the earthquake in all directions.

Explanation:

The vibrations produced by Earthquake are called seismic waves. seismic waves travel from the point where fault occurs. The maximum intensity is about the focus of the fault. These waves travel away from the focus in all directions.

Seismic waves are both transverse (S waves) and longitudinal (P waves). The P and S waves can travel through the Earth where as the surface waves travel above or near the Earth's surface.

kozerog [31]3 years ago
5 0

Answer: (C)

Explanation:  They travel away from the focus of the earthquake in all directions.

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Describe what would happen if you rubbed a mineral with a Mohs hardness value of 7 against a mineral with a value of 5?
chubhunter [2.5K]
The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.
8 0
3 years ago
Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?
denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

\lambda=\frac{2}{n} L


The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m


The wavelength of the 2nd harmonic is:

\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m


The wavelength of the 4th harmonic is:

\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m


It is not possible to find any integer n such that \lambda=5 m, therefore the correct options are A, B and D.

3 0
3 years ago
Read 2 more answers
You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in
ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f  = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5  + 0.6 × 20c = 343.5

so wavelength of sound  λ = c / f  

wavelength = 343.5 /  686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............   for n = 1, 2 3 .............

so

for d = 0.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1) = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2) = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3) = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475  = 2.975 m

distance 2 = 2.5 + 3.125  = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0
3 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
A wind turbine is an example of what kind of device
Liula [17]

Answer:

it’s an example of a generator.

Explanation:

3 0
2 years ago
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