Answer:
rate of heat exchange the animal and its environment. Feathers, vasoconstriction, Countercurrent heat exchanger and blubber are all mechanisms which restrict the transfer of heat from the animal body
Explanation:
Answer:
The change on the second particle is
.
Explanation:
The period of revolution of the particle in the magnetic field is given by the formula as follows :
![T=\dfrac{2\pi m}{Bq}](https://tex.z-dn.net/?f=T%3D%5Cdfrac%7B2%5Cpi%20m%7D%7BBq%7D)
It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.
![m_s=m_p](https://tex.z-dn.net/?f=m_s%3Dm_p)
If both particles take the same amount of time to go once around their respective circles. So,
![T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C](https://tex.z-dn.net/?f=T_e%3DT_s%5C%5C%5C%5C%5Cdfrac%7B2%5Cpi%20m_e%7D%7BBq_e%7D%3D%5Cdfrac%7B2%5Cpi%20m_s%7D%7BBq_s%7D%5C%5C%5C%5C%5Cdfrac%7Bm_e%7D%7Bq_e%7D%3D%5Cdfrac%7Bm_p%7D%7Bq_s%7D%5C%5C%5C%5Cq_s%3D%5Cdfrac%7Bm_pq_e%7D%7Bm_e%7D%5C%5C%5C%5Cq_s%3D%5Cdfrac%7B1.67%5Ctimes%2010%5E%7B-27%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B9.11%5Ctimes%2010%5E%7B-31%7D%7D%5C%5C%5C%5Cq_s%3D2.93%5Ctimes%2010%5E%7B-16%7D%5C%20C)
So, the change on the second particle is
.
El movimiento de las placas tectónicas
Answer:
(a) ![\frac{m}{s^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(b) ![\frac{m}{s^3}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%5E3%7D)
(c) 1 s
(d) 20 m
(e) 1 m
(f) ![0\frac{m}{s}](https://tex.z-dn.net/?f=0%5Cfrac%7Bm%7D%7Bs%7D)
(g) ![-12\frac{m}{s}](https://tex.z-dn.net/?f=-12%5Cfrac%7Bm%7D%7Bs%7D)
(h) ![-36\frac{m}{s}](https://tex.z-dn.net/?f=-36%5Cfrac%7Bm%7D%7Bs%7D)
(i) ![-72\frac{m}{s}](https://tex.z-dn.net/?f=-72%5Cfrac%7Bm%7D%7Bs%7D)
(j) ![-6\frac{m}{s^2}](https://tex.z-dn.net/?f=-6%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(k) ![-18\frac{m}{s^2}](https://tex.z-dn.net/?f=-18%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(l) ![-30\frac{m}{s^2}](https://tex.z-dn.net/?f=-30%5Cfrac%7Bm%7D%7Bs%5E2%7D)
(m) ![-42\frac{m}{s^2}](https://tex.z-dn.net/?f=-42%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Explanation:
Since <em>x</em> is measured in meters and <em>t</em> in seconds, constants <em>a </em>and <em>b</em> must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean
for <em>a </em>and
for <em>b</em>.
We can get the velocity <em>v </em>equation by deriving the position with respect to <em>t</em>, which gives:
![v=6*t-6*t^2](https://tex.z-dn.net/?f=v%3D6%2At-6%2At%5E2)
And the acceleration <em>a</em> equation by deriving again:
![a=6-12*t](https://tex.z-dn.net/?f=a%3D6-12%2At)
Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is <em>v=0</em>, which gives
![6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1](https://tex.z-dn.net/?f=6%2At-6%2At%5E2%3D0%5C%5C6t%2A%281-t%29%3D0%5C%5Ct%3D0%20or%20t%3D1)
For <em>t = 0</em>,<em> x = 0</em> so the maximun position is archieved at 1 second, which gives <em>x = 1 meter</em>.
For obtaining it's displacement <em>r</em>, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:
![r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B1%7D%7B4%7D%2A%282%2A4%5E3-3%2A4%5E2%29m%5C%5Cr%3D%2020m)
For the remaining questions, we just replace the values of <em>t</em> on the respective equations.