second question: How many seconds after the first snowball
should you throw the second so that they
arrive on target at the same time?
Answer in units of s.
Answer:
Part 1: 28°
Part 2: 1.367
Explanation:
Part 1:
Given: 62°
Simple
θ = 90°- 62°
<u>θ = 28°</u>
Part 2:
Y-direction
Δy
![t_{1} =\frac{2[16.2sin(62)]}{9.8}](https://tex.z-dn.net/?f=t_%7B1%7D%20%3D%5Cfrac%7B2%5B16.2sin%2862%29%5D%7D%7B9.8%7D)

![0=[16.2sin(28)]t_{2}+1/2(-9.8)t_{2}^{2}](https://tex.z-dn.net/?f=0%3D%5B16.2sin%2828%29%5Dt_%7B2%7D%2B1%2F2%28-9.8%29t_%7B2%7D%5E%7B2%7D)
![t_{2} =\frac{2[16.2sin(28)]}{9.8}](https://tex.z-dn.net/?f=t_%7B2%7D%20%3D%5Cfrac%7B2%5B16.2sin%2828%29%5D%7D%7B9.8%7D)

Δt
Δt
<u>Δt= 1.367s</u>
Hope it helps :)
wavelength =wavevelocity
--——————
Frequency
Frequency = 1/T => 1 / 6.73 = 0.1486
Wave velocity = L/T => 3.75 / 6.73 = 0.5572
Therefore, wave length = 0.5572/0.1486 = 3.75m
Answer:
The Earth is 4.543 billion years old
Answer:
D. 3.0 m/s
Explanation:
because I did this in my class
To solve this problem it is necessary to apply the concepts related to the Heisenberg's uncertainty principle. Under this principle we understand the relationship that the minimum range of error in position (x) times the minimum range of error in momentum (p) is, at a minimum, about equal to the Planck constant, mathematically that is,

Replacing with our values we have,


Therefore the least uncertainty in any simultaneous measurement of the momentum component px of this electron is 