Answer:
6.32m/s
Explanation:
note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²
okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!
when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)
now enough of the talking let solve....
so the ball was dropped .ie from rest to the ground through a distance of 2m,
the formula for calculating the distance if a body moving in a straight line is given by:
S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.
from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)
solving ....
we get t to be 0.632s
to find the speed we substitute t in the equation below:
V=u+at ,but since u=0
V=at =10•0.632=6.32m/s
therefore the speed the body uses to strike the ground is 6.32m/s
X - rays are highly penetrating electromagnetic radiations that are formed when the cathode rays strikes a dense metal.
It has many characteristics such as :
- travels with the speed of light.
Answer:
Neither lma0 I'm from a town :P
Explanation:
Hbu?
Have a nice dayyy <3
The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.
<h3>What is the law of conservation of momentum?</h3>
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
m is the mass =6.0 kg
t is the time interval=2 second
From Newton's second law;
From the graph;
The change in the momentum is;
Hence, the resulting change in momentum of the system will be +18.6 Ns.
To learn more about the law of conservation of momentum, refer;
brainly.com/question/1113396
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<span>it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m</span>
