I think its true i dont kno for sure
Their Period number is common means their "Principal Quantum Numbers" are same
Hope this helps!
Answer:
Total time taken=110 seconds
Total distance traveled=480m
Explanation:
First of all, we find the total time taken:
For that, we use the formula : Distance/Speed= Time
Time for part 1 : 200/5=40 seconds
Time for part 2 : 280/4=70seconds
Total time taken=110 seconds
Total distance traveled=480m
Average Speed= 480/110=4.36 m/s
Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)
Average Velocity=-80/110=-0.72 m/s
OR 0.72m/s towards west.
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
Given
Car 1
m1 = 1300 kg
v1 = 20 m/s
m2 = 900 kg
v2 = -15 m/s
(Negative sign shows that direction of car 2 is opposite to car 1)
Procedure
As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.