Answer:
Option B is correct.
A double insulated drill generally has a plastic case with the electrical connections and motor insulated within the tool
Explanation:
A double insulated drill's design in real life has plastic casing as the insulator & it wouldn't be a drill without its motor now, would it!?
Answer:
the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space
Explanation:
Answer:
a) 
b) 
Explanation:
a)
Given:
amount of heat transfer occurred,
initial temperature of car, 
final temperature of car, 
We know that the change in entropy is given by:

(heat is transferred into the system of car)

b)
amount of heat transfer form the system of house,
initial temperature of house, 
final temperature of house, 



Answer:
It is 52° below the celestial equator.
Explanation:
The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.
The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary
The star named Canopus has a declination of approximately –52°.
Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.
Thus, the star named Caponus is 52° below the celestial equator.
Explanation:
It is given that, the position of a particle as as function of time t is given by :

Let v is the velocity of the particle. Velocity of an object is given by :

![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)

So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :

![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)

At t = 0, 

Hence, this is the required solution.