Answer:
Q = 5267J
Explanation:
Specific heat capacity of copper (S) = 0.377 J/g·°C.
Q = MSΔT
ΔT = T2 - T1
ΔT=49.8 - 22.3 = 27.5C
Q = change in energy = ?
M = mass of substance =508g
Q = (508g) * (0.377 J/g·°C) * (27.5C)
Q= 5266.69J
Approximately, Q = 5267J
Answer choices ??? Or are there not any ?
Answer:
Part A:
(a): -121.26 ft/s.
(b): -121.13 ft/s.
(c): -121.052 ft/s.
(d): -121.026 ft/s.
Part B:
-121.00 ft/s.
Explanation:
Given that the height of the balloon after t seconds is
![\rm y(t) = 35 t-26t^2.](https://tex.z-dn.net/?f=%5Crm%20y%28t%29%20%3D%2035%20t-26t%5E2.)
The average velocity of an object is defined as the total distance traveled by the object divided by the time taken in covering that distance.
![\rm v_{av} = \dfrac{y_2-y_1}{t_2-t_1}](https://tex.z-dn.net/?f=%5Crm%20v_%7Bav%7D%20%3D%20%5Cdfrac%7By_2-y_1%7D%7Bt_2-t_1%7D)
where,
are the positions of the object at time
and
respectively.
<h2><u>
Part A:</u></h2><h2 />
- For the average velocity for the time period beginning when t=3 and lasting .01 sec.
For this case,
- For the average velocity for the time period beginning when t=3 and lasting .005 sec.
For this case,
![\rm t_1 = 3\ sec.\\t_2 = 3+0.005\ sec = 3.005\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.005-26\cdot 3.005^2=-129.60565\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.60565-(-129)}{3.005-3}=-121.13\ ft/s.](https://tex.z-dn.net/?f=%5Crm%20t_1%20%3D%203%5C%20sec.%5C%5Ct_2%20%3D%203%2B0.005%5C%20sec%20%3D%203.005%5C%20sec.%5C%5C%5C%5CTherefore%2C%20%5C%5C%5C%5Cy_1%20%3D%2035%5Ccdot%203-26%5Ccdot%203%5E2%3D-129%5C%20ft.%5C%5Cy_2%20%3D%2035%5Ccdot%203.005-26%5Ccdot%203.005%5E2%3D-129.60565%5C%20ft.%5C%5C%5C%5C%5CRightarrow%20v_%7Bav%7D%3D%5Cdfrac%7B-129.60565-%28-129%29%7D%7B3.005-3%7D%3D-121.13%5C%20ft%2Fs.)
- For the average velocity for the time period beginning when t=3 and lasting .002 sec.
For this case,
![\rm t_1 = 3\ sec.\\t_2 = 3+0.002\ sec = 3.002\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.002-26\cdot 3.002^2=-129.2421\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.2421-(-129)}{3.002-3}=-121.052\ ft/s.](https://tex.z-dn.net/?f=%5Crm%20t_1%20%3D%203%5C%20sec.%5C%5Ct_2%20%3D%203%2B0.002%5C%20sec%20%3D%203.002%5C%20sec.%5C%5C%5C%5CTherefore%2C%20%5C%5C%5C%5Cy_1%20%3D%2035%5Ccdot%203-26%5Ccdot%203%5E2%3D-129%5C%20ft.%5C%5Cy_2%20%3D%2035%5Ccdot%203.002-26%5Ccdot%203.002%5E2%3D-129.2421%5C%20ft.%5C%5C%5C%5C%5CRightarrow%20v_%7Bav%7D%3D%5Cdfrac%7B-129.2421-%28-129%29%7D%7B3.002-3%7D%3D-121.052%5C%20ft%2Fs.)
- For the average velocity for the time period beginning when t=3 and lasting .001 sec.
For this case,
![\rm t_1 = 3\ sec.\\t_2 = 3+0.001\ sec = 3.001\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.001-26\cdot 3.001^2=-129.121\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.121-(-129)}{3.001-3}=-121.026\ ft/s.](https://tex.z-dn.net/?f=%5Crm%20t_1%20%3D%203%5C%20sec.%5C%5Ct_2%20%3D%203%2B0.001%5C%20sec%20%3D%203.001%5C%20sec.%5C%5C%5C%5CTherefore%2C%20%5C%5C%5C%5Cy_1%20%3D%2035%5Ccdot%203-26%5Ccdot%203%5E2%3D-129%5C%20ft.%5C%5Cy_2%20%3D%2035%5Ccdot%203.001-26%5Ccdot%203.001%5E2%3D-129.121%5C%20ft.%5C%5C%5C%5C%5CRightarrow%20v_%7Bav%7D%3D%5Cdfrac%7B-129.121-%28-129%29%7D%7B3.001-3%7D%3D-121.026%5C%20ft%2Fs.)
<h2><u>
Part B:</u></h2>
The instantaneous velocity of the balloon at the given time is defined as the rate of change of its position at that time.
![\rm v(t) = \dfrac{dy}{dt}\\=\dfrac{d}{dt}\left ( 35 t-26t^2\right )\\\\=35-26\times 2t.\\\\At\ t=3,\\\\v(t)=35-26\times 2\times 3=-121.00\ ft/s.](https://tex.z-dn.net/?f=%5Crm%20v%28t%29%20%3D%20%5Cdfrac%7Bdy%7D%7Bdt%7D%5C%5C%3D%5Cdfrac%7Bd%7D%7Bdt%7D%5Cleft%20%28%2035%20t-26t%5E2%5Cright%20%29%5C%5C%5C%5C%3D35-26%5Ctimes%202t.%5C%5C%5C%5CAt%5C%20t%3D3%2C%5C%5C%5C%5Cv%28t%29%3D35-26%5Ctimes%202%5Ctimes%203%3D-121.00%5C%20ft%2Fs.)
<u>Note:</u><em> The negative sign with all the velocities indicates that the direction of these velocities are downwards.</em>
<em> </em>