Maunder minimum is related to climate due to the unusually low sunspot activity correlates to
unusually cold climatic events. The answer is letter A. It happened
around 1645 and 1715 and also coincided with the phenomena ‘Little Ice Age’
(1500 – 1850) in the Northern Hemisphere.
Answer:

Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by

Substitute the values then we get
Capacitance between parallel plate capacitor=

V=


Hence, the charge resides on the outer surface=
Answer:
An atom with a closed shell of valence electrons (corresponding to an electron configuration s2p6) tends to be chemically inert. An atom with one or two valence electrons more than a closed shell is highly reactive, because the extra valence electrons are easily removed to form a positive ion.Explanation:
Work = force x distance
= 100N (force) x 0.5m (distance)
= 50J
Explanation:
At point B, the velocity speed of the train is as follows.

= 
= 26.34 m/s
Now, we will calculate the first derivative of the equation of train.
y = 

Now, second derivative of the train is calculated as follows.
Radius of curvature of the train is as follows.
![\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7B%5B1%20%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdx%5E%7B2%7D%7D%7D)
= ![\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B1%20%2B%200.2e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B0.2%2810%5E%7B-3%7D%29e%5E%7B%5Cfrac%7B400%7D%7B1000%7D%7D%7D)
= 3808.96 m
Now, we will calculate the normal component of the train as follows.

= 
= 0.1822 
The magnitude of acceleration of train is calculated as follows.
a = 
= 
= 
Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is
.