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il63 [147K]
3 years ago
14

Which Richter magnitude range can be recorded by instruments but isn't felt? A. less than 2.9 B. 3.0 – 4.9 C. 5.0 – 5.9 D. 6.0 a

nd higher
Physics
2 answers:
Mashcka [7]3 years ago
6 0

Answer:

A. Less than 2.9

Explanation:

Richter scale is a measure of the intensity of the earthquakes designed and developed by Charles Richter in 1935. Its value ranges from 1.0 to 9.0 and greater. The earthquakes having a Richter magnitude of less than 2.9 are minor earthquakes which are slightly felt by a few humans but can be recorded on a seismograph. Earthquakes with magnitude above 2.9 can easily be felt by humans.

algol [13]3 years ago
3 0

A. anything less than 3.0 magnitude on a richters scale usually can't be felt by humans but instruments can pick it up.

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Find the wavelength of the third line in the lyman series, and identify the type of em radiation.
Natalija [7]

The wavelength of the third line in the Lyman series, and identify the type of EM radiation

In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.

1 / lambda = R(h)* ( \frac{1}{(n1)^{2} } -   \frac{1}{(n2)^{2} })

                 = 109678 ( \frac{1}{1^{2} } -  \frac{1}{3^{2} } )

                 = 109678 (8/9)

   Lambda = 9 / (109678 * 8 )

                  = 102.6 * 10^{-9} m = 102.6 nm

To learn more about Lyman series here

brainly.com/question/5762197

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8 0
1 year ago
A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25
solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

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3 0
3 years ago
What element are the cores of massive stars made of when a supernova occurs?
Studentka2010 [4]
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3 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
marta [7]
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
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7 0
3 years ago
Read 2 more answers
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