Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×
/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
Answer: Neon is a colorless gas at room temperature: Physical property
At room temperature, mercury is a liquid: Physical property
Apple slices turn brown when exposed to air: Chemical property
Phosphorus will ignite when exposed to air: Chemical property
Explanation:
Physical property is defined as the property of a substance which becomes evident during physical change in which there is alteration in shape, size etc. No new substance gets formed during physical change.
Example: Neon is a colorless gas at room temperature
At room temperature, mercury is a liquid.
Chemical property is defined as the property of a substance which becomes evident during chemical change in which a change in chemical composition takes place. A new substance is formed in these reactions.
Example: Apple slices turn brown when exposed to air: It undergoes oxidation which is a chemical change
Phosphorus will ignite when exposed to air: It undergoes oxidation which is a chemical change
The zeroth law of thermodynamics states that if two thermodynamic systems each are in thermal equilibrium with a third, then they are in thermal equilibrium with each other.
Displacement is zero. That’s because she ended up where she started.
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
