Answer: I actually need the same answer
Explanation:
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:
When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.
1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²
For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.
2. For this, you equate the y values of both balls:
y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds
Thus, the two balls would be at the same height after 2.25 seconds.
Answer:
388.97 nm
Explanation:
The computation of the wavelength of this light in benzene is shown below:
As we know that
n (water) = 1.333
n (benzene) = 1.501
And, the wavelength of water is 438 nm
![\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]](https://tex.z-dn.net/?f=%5Clambda%20%28benzene%29%20%3D%20%5Clambda%20%28water%29%20%5B%5Cfrac%7Bn%28water%29%7D%7Bn%28benzene%7D%5D)
Now placing these values to the above formula
So,
= 388.97 nm
We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come
Answer:
real, inverted, and smaller than the object
Explanation:
When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.
The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.
