<span>Niche - </span><span>An organism's particular role in an ecosystem, or how it makes its living.</span>
Answer:
![372.3 J/^{\circ}C](https://tex.z-dn.net/?f=372.3%20J%2F%5E%7B%5Ccirc%7DC)
Explanation:
First of all, we need to calculate the total energy supplied to the calorimeter.
We know that:
V = 3.6 V is the voltage applied
I = 2.6 A is the current
So, the power delivered is
![P=VI=(3.6)(2.6)=9.36 W](https://tex.z-dn.net/?f=P%3DVI%3D%283.6%29%282.6%29%3D9.36%20W)
Then, this power is delivered for a time of
t = 350 s
Therefore, the energy supplied is
![E=Pt=(9.36)(350)=3276 J](https://tex.z-dn.net/?f=E%3DPt%3D%289.36%29%28350%29%3D3276%20J)
Finally, the change in temperature of an object is related to the energy supplied by
![E=C\Delta T](https://tex.z-dn.net/?f=E%3DC%5CDelta%20T)
where in this problem:
E = 3276 J is the energy supplied
C is the heat capacity of the object
is the change in temperature
Solving for C, we find:
![C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C](https://tex.z-dn.net/?f=C%3D%5Cfrac%7BE%7D%7B%5CDelta%20T%7D%3D%5Cfrac%7B3276%7D%7B8.8%7D%3D372.3%20J%2F%5E%7B%5Ccirc%7DC)
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
Answer:
![T=51.64^\circ F](https://tex.z-dn.net/?f=T%3D51.64%5E%5Ccirc%20F)
![t=180.10s](https://tex.z-dn.net/?f=t%3D180.10s)
Explanation:
The Newton's law in this case is:
![T(t)=T_m+Ce^{kt}](https://tex.z-dn.net/?f=T%28t%29%3DT_m%2BCe%5E%7Bkt%7D)
Here,
is the air temperture, C and k are constants.
We have
in
So:
![T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F](https://tex.z-dn.net/?f=T%280%29%3D70%5E%5Ccirc%20F%5C%5CT%280%29%3D10%5E%5Ccirc%20F%2BCe%5E%7Bk%280%29%7D%5C%5C70%5E%5Ccirc%20F%3D10%5E%5Ccirc%20F%2BC%5C%5CC%3D70%5E%5Ccirc%20F-10%5E%5Ccirc%20F%3D60%5E%5Ccirc%20F)
And we have
in
, So:
![T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061](https://tex.z-dn.net/?f=T%2830%29%3D60%5E%5Ccirc%20F%5C%5CT%2830%29%3D10%5E%5Ccirc%20F%2B%2860%5E%5Ccirc%20F%29e%5E%7Bk%2830%29%7D%5C%5C60%5E%5Ccirc%20F%3D10%5E%5Ccirc%20F%2B%2860%5E%5Ccirc%20F%29e%5E%7Bk%2830%29%7D%5C%5C50%5E%5Ccirc%20F%3D%2860%5E%5Ccirc%20F%29e%5E%7Bk%2830%29%7D%5C%5Ce%5E%7Bk%2830%29%7D%3D%5Cfrac%7B50%5E%5Ccirc%20F%7D%7B60%5E%5Ccirc%20F%7D%5C%5C%2830%29k%3Dln%28%5Cfrac%7B50%7D%7B60%7D%29%5C%5Ck%3D%5Cfrac%7Bln%28%5Cfrac%7B50%7D%7B60%7D%29%7D%7B30%7D%3D-0.0061)
Now, we have:
![T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)](https://tex.z-dn.net/?f=T%3D10%5E%5Ccirc%20F%2B%2860%5E%5Ccirc%20F%29e%5E%7B-0.0061t%7D%281%29)
Applying (1) for
:
![T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F](https://tex.z-dn.net/?f=T%3D10%5E%5Ccirc%20F%2B%2860%5E%5Ccirc%20F%29e%5E%7B-0.0061%2A60%7D%5C%5CT%3D10%5E%5Ccirc%20F%2B%2860%5E%5Ccirc%20F%290.694%5C%5CT%3D10%5E%5Ccirc%20F%2B41.64%5E%5Ccirc%20F%5C%5CT%3D51.64%5E%5Ccirc%20F)
Applying (1) for
:
![30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s](https://tex.z-dn.net/?f=30%5E%5Ccirc%20F%3D10%5E%5Ccirc%20F%2B%2860%5E%5Ccirc%20F%29e%5E%7B-0.0061t%7D%5C%5C30%5E%5Ccirc%20F-10%5E%5Ccirc%20F%3D%2860%5E%5Ccirc%20F%29e%5E%7B-0.0061t%7D%5C%5C-0.0061t%3Dln%28%5Cfrac%7B20%7D%7B60%7D%29%5C%5Ct%3D%5Cfrac%7Bln%28%5Cfrac%7B20%7D%7B60%7D%29%7D%7B-0.0061%7D%3D180.10s)
Pressure is the amount of force exerted on an object and force is strength or energy of an action