We have
. So,
.
. So
.
Thus we can convert the units of the given quantity.
That is,
.
The quantity is converted to the required units.
Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
D-a is the answer can I get braillist
Answer:
E3 = 3.03 10⁻¹⁶ kJ, E4 = 4.09 10⁻¹⁶ kJ and E5 = 4.58 10⁻¹⁶ kJ
Explanation:
They give us some spectral lines of the Balmer series, let's take the opportunity to place the values in SI units
n = 3 λ = 656.3 nm = 656.3 10⁻⁹ m
n = 4 λ = 486.1 nm = 486.1 10⁻⁹ m
n = 5 λ=434.0 nm = 434.0 10⁻⁹ m
Let's use the Planck equation
E = h f
The speed of light equation
c = λ f
replace
E = h c /λ
Where h is the Planck constant that is worth 6.63 10⁻³⁴ J s and c is the speed of light that is worth 3 10⁸ m / s
Let's calculate the energies
E = 6.63 10⁻³⁴ 3 10⁸ / λ
E = 19.89 10⁻²⁶ /λ
n = 3
E3 = 19.89 10⁻²⁶ / 656.3 10⁻⁹
E3 = 3.03 10⁻¹⁹ J
1 kJ = 10³ J
E3 = 3.03 10⁻¹⁶ kJ
n = 4
E4 = 19.89 10⁻²⁶ /486.1 10⁻⁹
E4 = 4.09 10⁻¹⁹ J
E4 = 4.09 10⁻¹⁶ kJ
n = 5
E5 = 19.89 10⁻²⁶ /434.0 10⁻⁹
E5 = 4.58 10⁻¹⁹ J
E5 = 4.58 10⁻¹⁶ kJ