Answer:
601 nm
Explanation:
Energy of photon having wavelength of λ nm
= 
eV
Energy of 249 nm photon
=
=4.996 eV
Similarly energy of 425nm photon
=
=2.927 eV
Difference = 2.069 eV.
This energy will give rise to another photon whose wavelength will be
λ = 
= 601 nm.
Answer:
A) If one travels around a closed path adding the voltages for which one enters the negative reference and subtracting the voltages for which one enters the positive reference, the total is zero.
Explanation:
Kirchhoff's voltage law deals with the conservation of energy when the current flows in a closed-loop path.
It states that the algebraic sum of the voltages around any closed loop in a circuit is always zero.
In other words, the algebraic sum of all the potential differences through a loop must be equal to zero.
Answer:
the required height is 0.2449 m only
Explanation:
Given the data in the question;
Initial height = 2m
so speed of the puck before hitting the ground will be;
u² = 2gh
Initial speed u_ball = √2gh
u_ball = √( 2 × 9.8 × 2 )
u_ball = √39.2
u_ball = 6.26 m/s
given that; FOR THE FROZEN PUCK, coefficient of restitution = 0.35 only
R = - (v_ball - v_ground / u_ball - u_ ground)
so
0.35 = - (v_ball - 0 / 6.26 - 0)
0.35 = -v_ball / - 6.26
-v_ball = 0.35 × (- 6.26)
-v_ball = -2.191 m/s
v_ball = 2.191 m/s
to get the height;
v² = 2gh
h = v² / 2g
we substitute
h = (2.191)² / 2×9.8
h = 4.800481 / 19.6
h = 0.2449 m
Therefore, the required height is 0.2449 m only
The distance traveled while accelerating from rest is
D = 1/2 a t² .
For this problem, we shall totally ignore air resistance.
We do so completely without any reservation or guilt,
because we know that there is no air on the moon.
D = (1/2) · (1.6 m/s²) · (9 sec)²
= (0.8 m/s²) · (81 s²)
= (0.8 · 81) m
= 64.8 meters .
(That's about 213 feet ! The astronaut must have dropped the feather
from his spacecraft while he was aloft ... either just before touchdown
or just after liftoff.)
