Answer:
y = - 16t² + 55.6t + 6
Step-by-step explanation:
Using y - y₀ = vt - 1/2gt² where g = 32 ft/s², and v the velocity of the football
So y = y₀ + vt - 1/2 × (32 ft/s²)t²
y = y₀ + vt - 16t² where y₀ = 6.5 ft
y = 6 + vt - 16t²
Now, when t = 3.5 s, that is the time the teammate catches the ball after the quarterback throws it, y = 5 ft. Substituting these into the equation, we have
5 = 6.5 + v(3.5 s) - 16(3.5 s)²
5 = 6.5 + 3.5v - 196
collecting like terms, we have
5 - 6.5 + 196 = 3.5v
194.5 = 3.5v
v = 194.5/3.5 = 55.57 ft/s ≅ 55.6 ft/s
So, substituting v into y, our quadratic model is
y = 6 + 55.6t - 16t²
re-arranging, we have
y = - 16t² + 55.6t + 6
Answer:
if by distance around the hula hoop you mean circumference, then -
circumference = 2 pi r
= 2 X 22/7 X 14
= 4 X 22
= 88in
9514 1404 393
Answer:
(a) -- the correct choice is highlighted
Step-by-step explanation:
The units of specific heat tell you what quantities make up the ratio.
![\dfrac{390\text{ J}}{1\text{ kg$\cdot^\circ$C}}=\dfrac{-12.0\text{ J}}{0.012\text{ kg}\cdot\Delta T}\\\\\Delta T=\dfrac{-12.0}{0.012\cdot390}\ ^\circ\text{C}\approx-2.56\text{ $^\circ$C}](https://tex.z-dn.net/?f=%5Cdfrac%7B390%5Ctext%7B%20J%7D%7D%7B1%5Ctext%7B%20kg%24%5Ccdot%5E%5Ccirc%24C%7D%7D%3D%5Cdfrac%7B-12.0%5Ctext%7B%20J%7D%7D%7B0.012%5Ctext%7B%20kg%7D%5Ccdot%5CDelta%20T%7D%5C%5C%5C%5C%5CDelta%20T%3D%5Cdfrac%7B-12.0%7D%7B0.012%5Ccdot390%7D%5C%20%5E%5Ccirc%5Ctext%7BC%7D%5Capprox-2.56%5Ctext%7B%20%24%5E%5Ccirc%24C%7D)
The temperature will decrease by 2.56 C.