Answer:
2m/s^2
Explanation:
Clculate the acceleration:
V = u +at
20m/s = 0 + a*10s
a = 20m//10s
a = 2m/s²
From the data given , it is not possible to calculate the displacement , because no direction of motion is given
But it is possible to calculate the distance travelled
Distance = ut + ½ *a*t²
distance = 0 + ½ * 2m/s * 10²s
distance = 100m
Answer:
They will meet at a distance of 7.57 m
Given:
Initial velocity of policeman in the x- direction, 
The distance between the buildings, 
The building is lower by a height, h = 2.5 m
Solution:
Now,
When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.
Thus
From the second eqn of motion, we can write:
t = 0.707 s
Now,
When the policeman was chasing across:
The distance they will meet at:
9.57 - 2.0 = 7.57 m
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
Speed is the same as the initial: 25m/s.
*if* you need vectors though:
final velocity = (25*cos(35), -25*sin(35) ) m/s
