In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
By
vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by

where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by

where

is its direction with respect to the x-axis.
The water creates less friction between your foot and the ground
Answer:
Fundamental frequency in the string will be 25 Hz
Explanation:
We have given length of the string L = 1.2 m
Speed of the wave on the string v = 60 m/sec
We have to find the fundamental frequency
Fundamental frequency in the string is equal to
, here v is velocity on the string and L is the length of the string
So frequency will be equal to 
So fundamental frequency will be 25 Hz
Answer:
3400 m
Explanation:
Both lightning and thunder happen at the same time but one is faster than the other. The distance traveled by a sound can be calculated from its speed such that;
speed = distance/time, hence, distance = speed x time.
<em>For a thunder with 340 m/s speed and 10 seconds away from lightning, the distance between the thunder and the lightning can be calculated as</em>;
distance = 340 m/s x 10 s = 3400 m