Answer:
Velocity of Afrom B=21m/s
Acceleration of A from B=1.68m/s°2
Explanation:
Given
Radius r=150m
Velocity of a Va= 54km/hr
Va=54*1000/3600=15m/s
Velocity of b Vb=82km/hr
VB=81*1000/3600=22.5mls
The velocity of Car A as observed from B is VBA
VB= VA+VBA
Resolving the vector into X and Y components
For X component= 15cos60=7.5m/s
Y component=22 5sin60=19.48m/s
VBA= √(X^2+Y^2)
VBA= ✓(7.5^2+19.48^2)=21m/s
For acceleration of A observed from B
A=VA^2/r= 15^2/150=1.5m/s
Resolving into Xcomponent=1.5cos60=0.75m/s
Y component=3cos60=1.5
Acceleration BA=√(0.75^2+1.5^2)
1.68m/s
Answer: At 34°c
Explanation:
Using The Arrhenius Equation:
k = Ae − Ea/RT
k represents rate constant
A represents frequency factor and is constant
R represents gas constant which is = 8.31J/K/mol
Ea represents the activation energy
T represents the absolute temperature.
By taking the natural log of both sides,
ln k = ln A- Ea/RT
Reactions at temperatures T1 and T2 can be written as;
ln k1= ln A− Ea/RT1
ln k2= ln A− Ea/RT2
Therefore,
ln(k1/k2) = −Ea/RT1 + Ea/RT2
Since k2=2k1 this becomes:
ln(1/2) = Ea/R*[1/T2 − 1/T1]
Theefore,
-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]
1/T2 - 1/293 = -1.55 x 10^-4
1/T2 = -1.55 x 10^-4 + 34.13x 10^-4
1/T2 = 32.58 x 10^-4
Therefore T2 = 307K
T2 = 307 - 273 = 34 °c
Answer:
According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram(having magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex
Explanation:
2.392 hector liters is equal to 239.2 liters