It attracts two charged particles which are separated by a definitive distance.......
Answer:
12 m/s
Explanation:
Given data
Mass m= 1500kg
intitial velocity u= 20m/s
force F= 6000N
time t= 2 seconds
Required
The final velocity v
From
Ft= mΔv
Ft= m(u-v)
substitute
6000*2= 1500(20-v)
solve for v
12000= 30000- 1500v
collect like terms
12000-30000= -1500v
-18000= -1500v
divide both sides by -1500v
v= 18000/1500
v=12 m/s
Hence the velccity is 12 m/s
I'm not sure who plait is, but here's the problem with this plan that's so obvious most people miss it:
With one asteroid hurtling toward Earth, you have the possibility of one large body slamming into the Earth in one place.
Now, say we're able to successfully launch, guide, and detonate a big explosive weapon very close to the asteroid ... make it pull up alongside the asteroid, go BOOM, and shatter the asteroid into a thousand pieces.
THEN what do you have ?
You have a THOUSAND pieces, some of them quite large, with their center of mass still following the same trajectory that the original asteroid was. NOW there's a big fat chance that SEVERAL of these shattered pieces will hit the Earth ... maybe in a hundred DIFFERENT places.
NOW we're worse off than we were before the Orange One's Space Force decided that a nuclear weapon would be the answer to our looming problem.
My estimate is 1/2 inch per week.
(0.5 inch/wk) x (1 wk / 7 da) x (1 da / 86,400 sec) x (1 meter / 39.37 inches)
= (0.5 x 1 x 1 x 1) / (7 x 86,400 x 39.37) meter/second
= 1 / 47,621,952 meter/second
= 0.0000000209987... m/s
= 2.1 x 10⁻⁸ meter/second
= 2.1 x 10⁻⁵ mm/sec
= 0.000021 millimeter per second (rounded)
Answer:
Here, force=20N and displacement=10m
Work=Force×Displacement=20N×10m=200Nm
