Answer:
absorption and insolation.
Explanation:
At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?
Answer:
3.01L
Explanation:
Given parameters:
Initial pressure, P1 = 1.7atm
Initial volume, V1 = 4.25L
Final pressure, P2 = 2.4atm
Unknown:
Final or new volume, V2 = ?
Solution:
To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".
P1 V1 = P2 V2
P1 is the initial pressure
V1 is the initial volume
P2 final pressure
V2 final volume
1.7 x 4.25 = 2.4 x V2
V2 = 3.01L
Answer:
The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Explanation:
We are given that A city of Punjab has 15 percent chance of wet weather on any given day.
So, Probability of wet weather = 0.15
Probability of not being a wet weather = 1-0.15 =0.85
We are supposed to find probability that it will take a week for it three wet weather on 3 separate days
Total number of days in a week = 7
We will use binomial over here
n = 7
p =probability of failure = 0.15
q = probability of success=0.85
r=3
Formula :

Standard deviation =
Standard deviation =
Standard deviation =0.9447
Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Answer:
After 4 s of passing through the intersection, the train travels with 57.6 m/s
Solution:
As per the question:
Suppose the distance to the south of the crossing watching the east bound train be x = 70 m
Also, the east bound travels as a function of time and can be given as:
y(t) = 60t
Now,
To calculate the speed, z(t) of the train as it passes through the intersection:
Since, the road cross at right angles, thus by Pythagoras theorem:


Now, differentiate the above eqn w.r.t 't':


For t = 4 s:

Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .