To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.
Answer:
i think number 2 should be your pfp
Answer:3.56 nanosecond
In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:
time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft= 3.56
second or 3.56 nanosecond
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz
