Answer:
1. tropical storms and hurricanes
2.After a cold front moves through your area, you may notice that the temperature is cooler, the rain has stopped, and the cumulus clouds are replaced by stratus and stratocumulus clouds or clear skies.
3.Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface.
Explanation:
Answer:
option C is correct (250 g)
Explanation:
Given data:
Half life of carbon-14 = 5700 years
Total amount of sample = 1000 g
Sample left after 11,400 years = ?
Solution:
First of all we will calculate the number of half lives passes during 11,400 years.
Number of half lives = time elapsed/ half life
Number of half lives = 11,400 years/5700 years
Number of half lives = 2
Now we will calculate the amount left.
At time zero = 1000 g
At first half life = 1000 g/2 = 500 g
At second half life = 500 g/2 = 250 g
Thus, option C is correct.
Answer:
it causes alot of harm to the earth.
Explanation:
earthquakes
an so on
Answer:
5.31x10⁻⁶ C
Explanation:
The cube is located 100 m altitude from the ground, so the superior face is at 100m and has E = 70 N/C, and the inferior face is at the ground with E = 130 N/C.
The electric field is perpendicular to the bottom and the top of the cube, so the total flux is the flux at the superior face plus the flux at the inferior face:
Фtotal = Ф100m + Фground
Where Ф = E*A*cos(α). α is the angle between the area vector and the field (180° at the topo and 0° at the bottom):
Фtotal = E100*A*cos(180°) + Eground*A*cos(0°)
Фtotal = 70A*(-1) + 130*A*1
Фtotal = 60A
By Gauss' Law, the flux is:
Фtotal = q/ε, where q is the charge, and ε is the permittivity constant in vacuum = 8.854x10⁻¹² C²/N.m²
A = 100mx100m = 10000 m²
q = 60*10000*8.854x10⁻¹²
q = 5.31x10⁻⁶ C
Answer:
8.547 x 10⁴disintegrations per second
Explanation:
To calculate the disintegrations per second as -
Given ,
2.31 μCi of sulfur -35 .
Since ,
1 Ci = 3.7 * 10 ¹⁰ Bq
1 μCi = 10 ⁻⁶ Ci
Hence ,
conversation is done as follows -
2.31 ( 1 * 10⁻⁶) * ( 3.7 * 10¹⁰)
= 8.547 x 10⁴
Hence ,
8.547 x 10⁴disintegrations per second , the sample undergo for it to be brand new .
