Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
Answer:
The amount of heat transfer is 21,000J .
Explanation:
The equation form of thermodynamics is,
ΔQ=ΔU+W
Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.
Substitute 0 J for W and 0 J for ΔU
ΔQ = 0J+0J
ΔQ = 0J
The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change
The heat transfer is,
ΔQ=Q (in
) −Q (out
)
Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)
ΔQ=(19000J+2000J)−(0J)
=21,000J
Thus, the amount of heat transfer is 21,000J .
Answer:
The value of 
is 
.
Explanation:
Given that,
Number 
Energy difference = 6\times10^{-21}\ J[/tex]
Temperature T =300 K
We need to calculate the value of
We know that,
Hence, The value of is .
