<span>It can form four covalent bonds. </span>
Answer:
26 m/s
69 m
Explanation:
Given:
v₀ = 20 m/s
a = 2 m/s²
t = 3 s
Find: v and Δx
v = at + v₀
v = (2 m/s²) (3 s) + 20 m/s
v = 26 m/s
Δx = v₀ t + ½ at²
Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²
Δx = 69 m
Answer:
Final velocity will be 314.6 m/sec
Distance traveled = 1314.24 m
Explanation:
We have given initial velocity u = 233 m/sec
Acceleration 
Time t = 4.8 sec
From first equation of motion 
, here v is final velocity, u is initial velocity and t is time
So 
Now we have to find distance traveled
From second equation of motion
So distance traveled in given time will be 1314.24 m
Answer:
<em>The rebound speed of the mass 2m is v/2</em>
Explanation:
I will designate the two masses as body A and body B.
mass of body A = m
mass of body B = 2m
velocity of body A = v
velocity of body B = -v since they both move in opposite direction
final speed of mass A = 2v
final speed of body B = ?
The equation of conservation of momentum for this system is
mv - 2mv = -2mv + x
where x is the final momentum of the mass B
x = mv - 2mv + 2mv
x = mv
to get the speed, we divide the momentum by the mass of mass B
x/2m = v = mv/2m
speed of mass B = <em>v/2</em>
