Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
206Pb = 1.342 x10^22 atoms
<span>To find the number of atoms, you must first find the number of moles. If 238U is 238.029g/mol, and we have 1.75 grams, how many moles is that? 1.75 divided by 238.029 = 0.007352045 moles. To find the number of atoms in 0.007352045 moles, you multiply by a mole: </span>
<span>0.007352045 x 6.02 x 10^23 = 4.426 x10^21 atoms. </span>
<span>Same procedure for 206Pb: </span>
<span>4.59 divided by 205.97446 = 0.022284316 moles </span>
<span>0.022284316 x 6.02 x 10^23 = 1.342 x10^22 atoms. </span>
<span>Hope that helps you!
https://answers.yahoo.com/question/index?qid=20100331153014AAoMXcu
</span>
Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.
Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.
Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.
source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/
Bending occurs when one side of the wave enters the new medium before the other side of the wave. ... The bending occurs because the two sides of the wave are traveling at different speeds.
