Answer: student b ran up the stairs faster
Explanation:if they were going up the same stair case and student b got up there faster that means he was running therefore student b was using more power
The density of the substance is<u> 10.5 g/cm³.</u>
The jewelry is made out of <u>Silver.</u>
Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.
The volume <em>V</em> of the jewelry is given by,

Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

calculate the density ρ of the jewelry using the expression,

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

Since
,
The density of the jewelry is <u> 10.5 g/cm³.</u>
From standard tables, it can be seen that the substance used to make the jewelry is <u>silver</u><em><u>, </u></em>which has a density 10.5 g/cm³.
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:
ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)
Explanation:
Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :
ΔS al ≥ ∫dQ/T
if the heat transfer is carried out reversibly
ΔS al =∫dQ/T
in the surroundings
ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K
the total entropy change will be
ΔS total = ΔS al + ΔS surr
ΔS total ≥ ΔS al + (-ΔS al) =
ΔS total ≥ 0
the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings