Answer:
B.to examine a beam of charged particles
Explanation:
i took the quiz
Answer:
Explanation:
Given that on the tree the gravitational energy stored is 8J
Then, mgh = 8J.
The apple begins to fall and hit the ground, what is the maximum kinetic energy?
Using conservation of energy, as the above is about to hit the ground, the apple is at is maximum speed, and the height then is 0m, so the potential energy at the ground is zero, so all the potential of the apple at the too of the tree is converted to kinetic energy as it is about to hits the ground. Along the way to the ground, both the Kinetic energy and potential energy is conserved, it is notice that at the top of the tree, the apple has only potential energy since velocity is zero at top, and at the bottom of the tree the apple has only kinetic energy since potential energy is zero(height=0)
So,
K.E(max) = 8J
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
![0=[16.2sin(62)]t_{1}+1/2(-9.8)t_{1}^{2} \\](https://tex.z-dn.net/?f=0%3D%5B16.2sin%2862%29%5Dt_%7B1%7D%2B1%2F2%28-9.8%29t_%7B1%7D%5E%7B2%7D%20%5C%5C)
![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 :)
