Answer:Newton's three laws of motion relate to each other in that they lay a foundation for the principles of things in motion, then build upon that foundation. For example, the first law of motion,...
Explanation: WEEEEEEEEEEEEEEEEEEEEWOOOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWWWWWWWOOOOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWOOOOOOOOOOOOOOWWWWWWWWWWWWWWWWWWWWWWWWWWWWO-EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE
Answer: The following statement is true about squall line thunderstorm development: <em><u>These often form ahead of the advancing front but rarely behind it because lifting of warm, humid air and the generation of a squall line usually occur in the warm sector ahead of an advancing cold front. Behind a cold front, the air motions are usually downward, and the air is cooler and drier.</u></em>
<em>An upper-level wave, accountable for the fabrication of a squall line, extend in front of and backside a cold front, the air backside the front is cold, steady and settling while the air ahead of the front is hot and co-seismic.</em>
A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m
net work done by gravity on the cyclist = mass * gravity * height diff.
= 85 * 9.8 * 12.56
= 10470J
= 10.5kJ
B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J
vel.^2 = 10470 * 2 / 85 = 246.4
vel. = 15.7m/s
Answer:
True
Explanation:
If there's no preference over the string case (upper case or lower case), one can convert both strings to upper case or to lowercase and then compare the converted strings to test if they're equal or not.
An Illustration is
string a = "Boy"
string b = 'bOy"
if(a.ToUpper() == b.ToUpper() || a.ToLower() == b.ToLower())
{
Print "Equal Strings"
}
else
{
Print "Strings are not equal";
}
The above will first convert both strings and then compare.
Since they are the same (after conversion), the statement "Equal Strings" will be printed, without the quotes
For the answer to the question above, I assume that the question is two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3,?<span>two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3, is electrically neutral.
</span>From Gauss's law:
<span>Flux = ∫c E . dA = q/eo </span>
<span>Since this surface encloses all </span>
<span>charge, we can simplify: </span>
<span>Flux = (q1+q2+q3)/eo </span>
<span>Flux = </span>
<span>( (1*10^-6)+(-1.9*10^-6)+(0) )/(8.85*10^-12) = -101694.92 N·m2/C</span>
