Answer:
Explanation:
We shall apply the formula for velocity in case of elastic collision which is given below
v₁ = (m₁ - m₂)u₁ / (m₁ + m₂) + 2m₂u₂ / (m₁ + m₂)
m₁ and u₁ is mass and velocity of first object , m₂ and u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.
Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm
v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 ) + 2 x 18 x - 14 / ( 7.7 + 18 )
= - 8.817 - 19.6
= - 28.4 cm / s
K = 1/2 m x v^2
m = mass on the cart
V = velocity imparted to the cart
KA = 1/2 mA x vA^2.......................(1)
KB = 1/2 mB x vB^2........................(2)
Diving equation 1 by equation 2, we get -
KA/KB = mA/mB
= 2
KA = 2 x KB
Option A is correct
The magnitude of electric field is produced by the electrons at a certain distance.
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C
Answer:

Explanation:
It is given that,
Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :


R = Rydberg constant, 

Solving above equation we get the value of final n is,

or

So, it will relax in the n = 2. Hence, this is the required solution.
Explanation:
Below is an attachment containing the solution.