Deforestation, the chopping off of the trees that can take thousands or even millions of years to grow again<span />
Answer:
When they are connected in series
The 50 W bulb glow more than the 100 W bulb
Explanation:
From the question we are told that
The power rating of the first bulb is 
The power rating of the second bulb is 
Generally the power rating of the first bulb is mathematically represented as

Where
is the normal household voltage which is constant for both bulbs
So

substituting values

Thus the resistance of the second bulb would be evaluated as

From the above calculation we see that

This power rating of the first bulb can also be represented mathematically as

This power rating of the first bulb can also be represented mathematically as

Now given that they are connected in series which implies that the same current flow through them so

This means that

So when they are connected in series

This means that the 50 W bulb glows more than the 100 \ W bulb
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and

h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that


So on

= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
Acceleration will increase.
Explanation:
The relation between force, mass and acceleration according to the Newton's second law of motion is given as:
F = ma
We are given that the driving force on the truck remains constant, so F is constant here. We can rewrite the above equation as:

Since, F is constant, the acceleration of the truck is inversely proportional to the mass.
There is a hole at the bottom of the truck through which the sand is being lost at a constant rate. Since, the sand is being lost, the overall mass of the truck is being reduced.
Since, the acceleration of the truck is inversely proportional to the mass, the reduced mass will result in an increased acceleration.
So, the acceleration of the truck will increase.