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KatRina [158]
2 years ago
7

f an earthquake’s magnitude is measured based on the richter scale, how is this abbreviated? a. mr b. ms c. ml d. mb

Physics
1 answer:
zhannawk [14.2K]2 years ago
5 0
The answer is A. Hope this helps!
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Two parallel circular rings of radius R have their centres in the X axis separated by a distance L. If each ring carries a unifo
tatiyna

Answer:

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

Explanation:

The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x - axis is

E = Qx/4πε₀[√(x² + R²)]³

Since the rings are separated by a distance L, the electric field at point x due to the second ring is E' = -Q(L - x)/4πε₀[√((L - x)² + R²)]³. It is negative since it points in the negative x - direction.

So, the resultant electric field at x is E" = E + E' = Qx/4πε₀[√(x² + R²)]³ + {-Q(L - x)/4πε₀[√((L - x)² + R²)]³}

E" =  Qx/4πε₀√[(x² + R²)]³ - Q(L - x)/4πε₀√[((L - x)² + R²)]³

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[((L - x)² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L² - 2Lx + x² + R²)]³})

E" =  Q/4πε₀(x/√[(x² + R²)]³ - (L - x)/√[(L - 2x)L + (x² + R²)]³})

E" =  Q/4πε₀√[(x² + R²)]³(x - (L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

So, the electric field at points along the x axis is

E" =  Q/4πε₀√[(x² + R²)]³(x - {(L - x)/√[(L - 2x)L/(x² + R²) + 1]³})

5 0
3 years ago
QUESTION 5
Fiesta28 [93]

Answer:

D) sudden and irresistible on sets of sleep during normal waking periods

Explanation:

D because Narcolepsy (sudden and irresistible on sets of sleep during normal waking periods) is a part of insomnia

8 0
3 years ago
Metals in group 2 on the period table most commonly form which type
LuckyWell [14K]

Answer:

alkaline earth metals

Group 2 metals, the alkaline earth metals, have 2 valence electrons, and thus form M2+ ions. The halogens, Group 17 , reach a full valence shell upon reduction, and thus form X− ions

Explanation:

6 0
3 years ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
If an object goes through a
aivan3 [116]

Answer:

A, The same amount of gravity

Explanation:

4 0
2 years ago
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