Answer:
The appropriate solution is:
(a) 
(b) 
(c) 
Explanation:
According to the question, the value is:
Power of bulb,
= 60 W
Distance,
= 1.0 mm
Now,
(a)
⇒ 
On applying cross-multiplication, we get
⇒ 
⇒ 
⇒ 
(b)
As we know,
⇒ 
By putting the values, we get
⇒ 
(c)
⇒ 

⇒ 
⇒ 
The weather report never tells the "humidity" of the air. It tells the
"RELATIVE humidity". That's the percent of the moisture the air
COULD hold at the current temperature that it's actually holding
right now.
If the air is completely dry, then it's holding NONE of the moisture
that it COULD hold, and the relative humidity is zero percent.
Answer:
The charge density in the system is 
Explanation:
To solve this problem it is necessary to keep in mind the concepts related to current and voltage through the density of electrons in a given area, considering their respective charge.
Our data given correspond to:

We need to asume here the number of free electrons in a copper conductor, at which is generally of 
The equation to find the current is

Where
I =Current
V=Velocity
A = Cross-Section Area
e= Charge for a electron
n= Number of free electrons
Then replacing,


Now to find the linear charge density, we know that

Where:
I: current intensity
Q: total electric charges
t: time in which electrical charges circulate through the conductor
And also that the velocity is given in proportion with length and time,

The charge density is defined as

Replacing our values


Therefore the charge density in the system is 
Answer:
(A) 
(B) s = 146.664 m
Explanation:
We have given car starts from the rest so initial velocity u = 0 m /sec
Final velocity v = 88 km/hr
We know that 1 km = 1000 m
And 1 hour = 3600 sec
So 
Time is given t = 12 sec
(A) From first equation of motion v = u+at
So 

So acceleration of the car will be 
(b) From third equation of motion 
So 
s = 146.664 m
Distance traveled by the car in this interval will be 146.664 m
Answer:
Explanation:
efficiency of carnot engine operating between 1100 K and 320 K
= 1100 - 320 / 1100
= .709
efficiency of heat engine = 120 / 400 = .30
2 nd law thermal efficiency of heat engine = .30 x 100 / .709
= 42.31 % .