Answer:
I need this for may schooling
Answer:
4.9 x 10^-19 J, 2.7 x 10^-19 J
Explanation:
first wavelength, λ1 = 410 nm = 410 x 10^-9 m
Second wavelength, λ2 = 750 nm = 750 x 10^-9 m
The relation between the energy and the wavelength is given by
E = h c / λ
Where, h is the Plank's constant and c be the velocity of light.
h = 6.63 x 10^-34 Js
c = 3 x 10^8 m/s
So, energy correspond to first wavelength
E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J
E1 = 4.9 x 10^-19 J
So, energy correspond to second wavelength
E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J
E2 = 2.7 x 10^-19 J
Answer:
15.3 s and 332 m
Explanation:
With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon
gm = 1/6 ge
gm = 1/6 9.8 m/s² = 1.63 m/s²
We calculate the range
R = Vo² sin 2θ / g
R = 25² sin (2 30) / 1.63
R= 332 m
We will calculate the time of flight,
Y = Voy t – ½ g t2
Voy = Vo sin θ
When the ball reaches the end point has the same initial height Y=0
0 = Vo sin t – ½ g t2
0 = 25 sin (30) t – ½ 1.63 t2
0= 12.5 t – 0.815 t2
We solve the equation
0= t ( 12.5 -0.815 t)
t=0 s
t= 15.3 s
The value of zero corresponds to the departure point and the flight time is 15.3 s
Let's calculate the reach on earth
R2 = 25² sin (2 30) / 9.8
R2 = 55.2 m
R/R2 = 332/55.2
R/R2 = 6
Therefore the ball travels a distance six times greater on the moon than on Earth
Answer:
Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.
Explanation:
Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
Hope this helps!
