Answer:
0.906
Explanation:
Let g = 9.81 m/s2. We can calculate the rate of change in potential energy when m = 201kg of water is falling down a distance of h = 131m per second
So the efficiency of the water turbine is the ratio of output power over input power:
Answer:
Acceleration will be equal to 
Explanation:
We have given mass of the object m = 0.4 kg
Spring constant k = 8 N/m
Maximum displacement of the spring is given x = 0.1 m
From newton's law force is equal to 
.....eqn 1
By hook's law spring force is equal to 
.....eqn 2
From equation 1 and equation 2
So acceleration will be equal to
Answer:
817.5 Pa
Explanation:
From Bernoulli's equation, considering thst there is no height difference then
P1+½d(v1)²=P2+½d(v2)²
P1-P2=½d(v2²-v1²)
∆P=½d(v2²-v1²)
Where P represent pressure, d is density and v is velocity. Subscripts 1 and 2 represent inside and outside. ∆P is tge change in pressure
Given the speed at roof top as 128 km/h, we convert it to m/s as follows
128*1000/3600=35.555555555555=35.56 m/s
Velocity at the bottom of roof is 0 m/s
Density is given as 1.293 kg/m³
∆P=½*1.293*(35.56²-0)=817.5 Pa
Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
And so
And using
, we find K2:
