The separation in time between the arrival of primary and secondary wave is called LAG TIME.
The time difference between the arrival of primary wave and secondary wave in a seismogram is called lag time. The primary wave always travels faster than the secondary wave, thus the difference between the two can be obtained by estimating the difference between the arrival time of the two waves/.
Answer:
he polarity of the electromagnet is determined by the direction the current. The north pole of the electromagnet is determined by using your right hand. Wrap your fingers around the coil in the same direction as the current is flowing (conventional current flows from + to -).
Explanation:
Answer:
age of the site is 15411.75 years old
Explanation:
Given data
plant or animal dies = 14C
time period = 5730 year
carbon = 15.5%
to find out
age (in years) of the ancient site
solution
we know that Final value = Initial value × 
here n is half life passed
so for 15.5%
15.5% = 100% of
0.155 = 1 ×
now take log both side
log 0.155 = log
n = log 0.155 / log 0.5
n = 2.68966
we know here 5730 years in half life
so for 2.68966 half-lives = 2.68966 × 5730 = 15411.7518
age of the site is 15411.75 years old
Answer;
Amount of time the two substances are in contact
Area in contact between the two substances
Specific heat of the material that makes up the substances
Explanation;
The change in temperature of a substance is caused by heat energy. The change in temperature will depend on factors such as mass of the substance, the type of material it is made from, the time taken , specific heat of the material that makes the substance, and also the area of contact.
The amount of time the two substances are in contact affect the change in temperature such that if the two bodies are in contact for a longer time then a bigger change in temperature will be observed.
Specific heat capacity also determines the change in temperature that will be observed, such that a substance with a bigger specific heat capacity will record a small change in temperature.
Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
