1) n=1 -> n=2 : delta E = -5+11 = 6. The answer is D
<span>2) n=1 -> n=3 : delta E = -2+11 = 9. The answer is B </span>
<span>3) n=1 -> n=4 : delta E = -1+11 = 10. No solution available </span>
<span>4) n=1 -> infinity delta E = 11. The answer is A </span>
<span>5) not absorbed would be C, as there is no transition with delta E of 8. </span>
Answer:
option B
Explanation:
given,
height of building = 0.1 km
ball strikes horizontally to ground at = 65 m
speed at which the ball strike = ?
vertical velocity = 0 m/s
time at which the ball strike
t = 4.53 s
vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s
horizontal velocity = 
=14.35 m/s
speed of the ball = 
= 46.65 m/s
hence, the speed of the ball just before it strike the ground = 47 m/s
The correct answer is option B
Answer:
D
Explanation:
<em>The most suitable testable question. in this case, would be that 'are there more home runs during the more humid months of the summer?'</em>
Since the aim of the investigation is to find the relationship between humidity and the number of home runs, measuring the number of home runs during the more humid months in the summer and comparing the data to the number of home runs during the less humid months in the same summer would provide the answer.
<u>Only option D raises a valid question that is relevant to the aim of the investigation.</u>
The other 4 kg of mass may have departed the scene
of the fire, in the form of gases and smoke particles.
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
