A great, helpful, useful definition of acceleration is
<em>A = (change in speed) / (time for the change)</em> . <== you should memorize this
This simple tool will directly solve all 3 problems.
The REASON for assigning these problems for homework is NOT to find the answers. It's to help YOU find out whether you know this definition, to let you go back and review it if you don't, and to give you a chance to practice using it if you do. Noticed that if you get the answers from somebody else, you lose all of these benefits.
The only wrinkle anywhere here is in #3, because when you use this definition, the unit of time has to be the same in both the numerator and the denominator.
So for #3, you have to EITHER change the km/hr to km/sec, OR change the 4sec to a fraction of an hour, before you plug anything into the definition.
Answer:

Explanation:
Let the distance from spotlight to wall be 15m, and distance from the man to the building be
.
#Therefore the height of the shadow as a function of the above is 
Hence, height of the shadow is expressed as s=(15-x)m
#See attached photo for illustration
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4
= 
So,
Rearranging the above equation to get Electric field, we will get:
E = 
Multiply and divide by
E =
x 
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x
) /(
x
)
c) r > r2 :
Electric Field = ?
E x 4
= 
Rearranging the above equation for E:
E = 
E =
+ 
As we know from above, that:
= (σ1 x
) /(
x
)
Then, Similarly,
= (σ2 x
) /(
x
)
So,
E =
+ 
Replacing the above equations to get E:
E = (σ1 x
) /(
x
) + (σ2 x
) /(
x
)
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x
= - σ2 x 
Rain fall which is called run off
Answer:
B, C and E
Explanation:
The unit of resistance in the international system is the Ohm, the equation that describes the resistance is:

Where (l) is for lenght of the wire, (S) is the area and (p) its the constant associated to the conductor.
It's related by the Ohm's Law:
