The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.
<h3>How we compare the flow of electricity to water?</h3>
Water flowing in pipes is like flowing of electricity in a circuit. A battery is like a pump from where electricity comes and moves in the circuit. Electrons flowing through wires are like water molecules flowing through pipes. So in comparison between water and electricity, both water and electricity are similar to each other in flowing and movement.
So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.
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Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
Answer:
the sides of the wedge are inclined.
Explanation:
The wedge is a triangular simple machine with a blunt face and two inclined faces. The distribution of forces in a wedge is because the sides of the wedge are inclined.
Answer:
T2=336K
Explanation:
Clausius-Clapeyron equation is used to determine the vapour pressure at different temperatures:
where:
In(P2/P1) = ΔvapH/R(1/T1 - 1/T2)
p1 and p2 are the vapour pressures at temperatures
T1 and T2
ΔvapH = the enthalpy of vaporization of the liquid
R = the Universal Gas Constant
p1=p1, T1=307K
p2=3.50p1; T2=?
ΔvapH=37.51kJ/mol=37510J/mol
R=8.314J.K^-1moL^-1
In(3.50P1/P1)= (37510J/mol)/(8.314J.K^-1)*(1/307 - 1/T2)
P1 and P1 cancelled out:
In(3.50)=4511.667(T2 - 307/307T2)
1.253=14.696(T2 - 307/T2)
1.253=(14.696T2) - (14.696*307)/T2
1.253T2=14.696T2 - 4511.672
Therefore,
4511.672=14.696T2 - 1.253T2
4511.672=13.443T2
So therefore, T2=4511.672/13.443=335.61
Approximately, T2=336K
