Much of the precipitation in large bodies of water occurs at the surface. The ocean loses about 37000 km cubed considering evaporation and precipitation.
Answer:
0.000000002 m=2.0*10⁻⁹ m
Explanation:
Scientific notation allows us to write very large or very small numbers in abbreviated form. This notation simply consists of multiplying by a power of base 10 with a positive or negative exponent.
A number written in scientific notation has the form:
a*10ⁿ
where:
- the coefficient a has a value such that 1 ≤ a <10
- n is an integer. Represents the number of times the decimal point is shifted. It is always a whole number, positive if it is shifted to the left, negative if it is shifted to the right.
So to write the number 0.000000002 in scientific notation, the following steps are performed:
- The decimal point is moved to the right as many spaces until it reaches the right of the first digit.
- This number is then written, which will be the coefficient a in the expression of the previous product. So a=2.0
- The base 10 is written with the exponent equal to the number of spaces that the comma moves. So n=9. But this is a negative number because the comma shifts to the right.
So, you get: <u><em>0.000000002 m=2.0*10⁻⁹ m</em></u>
Answer:
The element "AI" has:
Protons: 13 Neutrons: 14 Electrons: 13
Have a great day.
Answer:
Explanation:
The volume of contaminated water
= cross sectional area x height of water level
3.14 x 9 x 9 x 7.5 ft³
= 1907.55 ft³
mass = density x volume
= 1907.55 x 63.5 lbs
m = 121129.425 lbs
This mass has to be raised to the height of 8 ft before evacuation .
There is a rise of centre of mass of
8 - 7.5/2 ft
h = 4.25 ft
Energy required
= mgh
= 121129.425 x 32 x 4.25
= 16473601.8 unit.
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :



Work done on the wagon is given by the product of net force and displacement. It is given by :


W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.